3. Stoichiometry questions

Michael Mombourquette

Balance the following equations:
1. C₂H₅OH + O₂ → CO₂ + H₂O
2. BCl₃ + H₂O → B(OH)₃ + HCl
3. NiS + O₂ → NiO + SO₂
4. NH₄VO₃ → V₂O₅ + NH₃ + H₂O
5. N₂O₅ + H₂O → HNO₃
6. Ca₃(PO₄)₂ + SIO₂ + C → P₄ + CaSiO₃ + CO
Methane burns with oxygen according to the equation
CH4 + 2O₂ → CO₂ + 2H₂O
1. How many moles of oxygen are required per mole of methane?
2. How many kilograms of oxygen are required per kilogram of methane? [2, 3.99]
A sample of a mixture of barium chloride salt and sodium chloride salt was completely dissolved in nitric acid. Upon adding dilute sulfuric acid to the solution, virtually all the barium was precipitated as white BaSO₄, which was filtered, dried and weighed. If the mass of the original sample was .2270 g and the mass of the precipitate was 0.0235 g, calculate the mass fraction of barium in the mixture. [0.0922]
Equal masses of sodium metal and water are reacted to form hydrogen gas, H₂, and sodium hydroxide, NaOH. Write a balanced chemical equation for the reaction. Which of the reactants is in excess if the reaction goes to completion? what fraction of the reactant remains unreacted? [H₂O,.216]
Baking powder is a mixture of sodium bicarbonate and a solid state acid. When dissolved in water, carbon dioxide gas is released. Some examples of the reactions found in different brands of baking powders are shown below:
KH₂PO₄ + NaHCO₃ → CO₂ + KNaHPO₄ + H₂O
Na₂Al₂(SO₄)₄ + 6 NaHCO₃ → 6 CO₂ + 4 Na₂SO₄ + 2Al(OH)₃
If each of the powders represented by these reactions consists of a stoichiometric mixture (meaning all reactants are exactly used up together), calculate the amount of CO₂ (in moles) which would be released by 1.00 g of each kind of baking powder. [4.5×10^-3 mol, 6.07×10^-3 mol]
A sample of seawater is titrated with an aqueous solution of silver nitrate, taking advantage of the reaction in which silver ion reacts with chloride ion to precipitate insoluble silver chloride (AgCl). What is the concentration of chloride ion in the seawater if it takes 19.7 mL of 0.100 M AgNO3 solution to precipitate all the chloride ion from a 10.0 mL sample of seawater? What is the mass of AgCl precipitated? [0.197 mol/L, 0.282 g]
A sample of a mixture of KBr and NaBr weighing 0.5605 g was treated with AgNO3 and all the bromide ion was precipitated as AgBr(s). When dried, the AgBr weighed 0.9702g. What was the mass percent of the KBr in the original sample?[38]
What is the volume of a 0.563 M solution of NaOH that is required to react completely with 10.0 mL of a 0.255 M solution of HNO₃? [4.53 mL]
What volume of a 0.563 M solution of NaOH is required to react completely with 10.0 mL of a .255 M solution of acetic acid? [4.53 mL]
What volume of a 0.563 M solution of NaOH is required to react completely with 10.0 mL of a .255 M solution of H₂SO₄? [9.06 mL]

Answers

Balance the following equations:
1. C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O
2. BCl₃ + 3H₂O → B(OH)₃ + 3HCl
3. NiS + 1 1/2O₂ → NiO + SO₂ or 2NiS + 3O₂ → 2NiO + 2SO₂
4. 2NH₄VO₃ → V₂O₅ + 2NH₃ + H₂O
5. N₂O₅ + H₂O → 2HNO₃
6. 4Ca₃(PO₄)₂ + 12SIO₂ + 3C → P₄ + 12CaSiO₃ + 4CO
Methane burns with oxygen according to the equation
CH₄ + 2O₂ → CO₂ + 2H₂O
1. How many moles of oxygen are required per mole of methane? The stoichiometric ratio (from the balanced equation) gives 2 moles of O₂ for every 1 mole of CH₄.
2. How many kilograms of oxygen are required per kilogram of methane?
  Molar mass of CH₄ is $12.01 + 4\times 1.008 = 16.04\frac{g}{mol}$ .
  Molar mass of oxygen gas is $2\times16.01 = 32.02\frac{g}{mol}$ .
  now use the molar mass in the calculation along with the stoichiometric mole ratios to get the answer. $\frac{1 kg\; CH_4}{}\times \frac{1\; mol}{16.04\; g}\times \frac{2 O_2}{1 CH_4}\times\frac{32.02 g}{1 mol} = 3.99\;g\;O_2$
A sample of a mixture of barium chloride salts and sodium chloride salts was completely dissolved in nitric acid. Upon adding dilute sulfuric acid to the solution, virtually all the barium was precipitated as white BaSO₄, which was filtered, dried and weighed. If the mass of the original sample was .2270 g and the mass of the precipitate was 0.0235 g, calculate the mass fraction of barium in the mixture. [0.0922]
First, write the reaction between the barium cation and sulfuric acid that made the solid BaSO₄. Note that the barium salt consists of a Ba²⁺cation and 2 Cl^– anions with a formula of BaCl₂. Similarly, the sodium chloride is NaCl. The molar mass of barium chloride is 207.843 g/mol. we won’t need the molar mass for sodium chloride.
Ba²⁺ + SO₄^2- → BaSO₄. So, it’s a 1:1 mole ratio of the barium salt to the barium sulfate. we know the mass of the precipitate, so we can get the moles of the precipitate using the molar mass of BaSO₄.
$0.0235\;g BaSO_4\times\frac{1 mol}{233.857 g}\times \frac{1\;BaCl_2}{1\;BaSO_4}\times\frac{207.843 g}{mol}= 0.0209 g BaCl_2$
Now, divide this by original amount of sample to get the mass ratio $\frac{.0209}{.2270} =0.0922$ Clearly, most of the original sample was sodium chloride and only a tiny amount was the barium salt.
Equal masses of sodium metal and water are reacted to form hydrogen gas, H₂, and sodium hydroxide, NaOH. Write a balanced chemical equation for the reaction. Which of the reactants is in excess if the reaction goes to completion? what fraction of the reactant remains unreacted? [H₂O,.216]
First, write the balanced equation for the reaction of sodium metal with water to form hydrogen gas.
Na(s) + H₂O(l) → NaOH + 1/2 H₂. or, using whole numbers,
2Na(s) + 2H₂O(l) → 2NaOH + H₂.
So, there is a 1:1 mole ratio of Na(s) to water. let’s pick 1 mole of each and calculate the mass. Na: 22.9898g H₂O: 18.015 g. the question said equal masses were added but if 18.015 g of water is there, then it needed 22.989 g of Na but there was only 18.015 g so Na is limiting and H₂O is in excess. so all the sodium is used up first and the remainder (water) is
$18.015\;g\;Na\times\frac{1 mol Na}{22.9898\;g\;Na}\times\frac{1 H_2O}{1Na}\times\frac{18.015\;g\;H_2O}{1mol\;H_2O} = 14.116\;g\;H_2O$
But, we had $18.015 g H_2O$ so there is $18.015-14.116 = 3.9 g$ of water left over. The fraction left over is $\frac{3.9}{18.015} = 0.216$
Baking powder is a mixture of sodium bicarbonate and a solid state acid. When dissolved in water, carbon dioxide gas is released. Some examples of the reactions found in different brands of baking powders are shown below:
KH₂PO₄ + NaHCO₃ → CO₂ + KNaHPO₄ + H₂O
Na₂Al₂(SO₄)₄ + 6 NaHCO₃ → 6 CO₂ + 4 Na₂SO₄ + 2Al(OH)₃
If each of the powders represented by these reactions consists of a stoichiometric mixture (meaning all reactants are exactly used up together), calculate the amount of CO₂ (in moles) which would be released by 1.00 g of each kind of baking powder. [4.5×10^-3 mol, 6.07×10^-3 mol]
Let’s do each brand of baking powder in turn.
$M(KH_2PO_4) = 136.085 g/mol \;\; M(NaHCO_3)=84.007g/mol$
We have one gram of the complete power so
$m(KH_2PO_4)+m(NaHCO_3)=1.00g$
but we don’t know either mass so we need more equations (simultaneous linear equations). We are told the mixture is stoichiometric so that means
$n(KH_2PO_4)=n(NaHCO_3)$ .
Still need more equations to solve this. We know how to replace the m in the equations by n using the molar masses $m=n\times M$ , so let’s change out the first equation and replace the masses with molar masses, which we know.
$n(KH_2PO_4)\times 136.085g/mol + n(NaHCO_3)\times 84.007g/mol=1.00g$
since we know that both number of moles are the same, we can just use n and solve for n.
$n\times 136.085g/mol + n\times 84.007g/mol=1.00g$
$n = 0.00454 mol$
now, using the 1:1 stoichiometry,
$n(CO_2) = 4.5\times10^{-3}\;moles$ .
OK, Let’s repeat that process for the second equation
$M(Na_2Al_2(SO_4)_4) = 484.193 g/mol ;\;\; M(NaHCO_3)=84.007g/mol$
$m(Na_2Al_2(SO_4)_4)+m(NaHCO_3)=1.00g$
$6\times n(Na_2Al_2(SO_4)_4)=n(NaHCO_3)$
This time, the stoichiometry is not 1:1, so let n be the number of moles of $Na_2Al_2(SO_4)_4$ .
$n(Na_2Al_2(SO_4)_4)\times484.193 g/mol+6n(NaHCO_3)\times84.007g/mol=1g$
$n\times 484.193g/mol + 6\times n\times 84.007g/mol=1.00g$
$n(Na_2Al_2(SO_4)_4) = 0.00101 mol$
now, using the 1:6 stoichiometry,
$n(CO_2) = 6\times n(Na_2Al_2(SO_4)_4) = 6\times 0.00101 mol = 6.06\times10^{-3}\;moles$
A sample of seawater is titrated with an aqueous solution of silver nitrate, taking advantage of the reaction in which silver ion reacts with chloride ion to precipitate insoluble silver chloride (AgCl). What is the concentration of chloride ion in the seawater if it takes 19.7 mL of 0.100 M AgNO3 solution to precipitate all the chloride ion from a 10.0 mL sample of seawater? What is the mass of AgCl precipitated? [0.197 mmol/L, 0.282 g]
First, write the balanced equation between the silver nitrate and the chloride ions from seawater.
Cl^–(aq)+AgNO₃(aq) → AgCl(s) + NO₃^–(aq) . Since the nitrate ions are soluble with everything, we can just leave them as ions dissolved in water.
So, now we find the number of moles of silver nitrate we need and from there, we can use the stoichiometry to get the number of moles of chloride in the 10.0 mL of seawater.
$19.7mL\;AgNO_3\times 0.100\;\frac{mol}{L}=1.97\; mmol AgNO_3 \;or\; 1.97\times10^{-3}\;moles$
So, we now know the moles of silver nitrate and according to the stoichiometry, that is equal to the moles of chloride in the 10.0 mL of seawater. Let’s now find the concentration of the seawater.
$\underline{M}(Cl^- \;in \;seawater)=\frac{1.97\; mmol}{10.0\;mL}=0.197\frac{mol}{L}$ Note: the prefix mili cancels out, underlined M means concentration in moles per litre
We also, by stoichiometry know the moles of silver nitrate precipitated, so the mass can be calculated using the molar mass of AgCl : $M(AgCl)=143.32\frac{g}{mol}$
$m(AgCl)=0.197\;mmol\times143.32\frac{g}{mol} = 0.282 mg$
A sample of a mixture of KBr and NaBr weighing 0.5605 g was treated with AgNO3 and all the bromide ion was precipitated as AgBr(s). When dried, the AgBr weighed 0.9702g. What was the mass percent of the KBr in the original sample?[38]
Mass percent is the mass in g that would occur in 100g, written as %m/m, which means “percent by mass, aka, per 100, mass over mass”.
So, we need to find the mass of the KBr and then divide by total mass (gives us the mass ratio) and then multiply by 100% (100/100).
$0.9702\;g\;AgBr\times \frac{1\; mol}{187.77\;g}=0.005167 mol\;AgBr.$
We know only the total mass of the mixture:
1 $m(KBr)+m(NaBr)=0.5605\;g$
We know the total number of moles of Br, and hence, the total of the number of moles of KBr and NaBr or just the moles of K and Na.
2 $n(KBr)+n(NaBr)=.005167\;mol$
Rewrite equation 1 using the molar masses to get rid of the mass of each salt. $M(KBr)=119.002g/mol\;\;\; M(NaBr)=102.894g/mol$
3. $119.002\times n(KBr) + 102.89\times n(NaBr) = 0.5605g$
Now we have two equations (2,3) in two unknowns $(n(NaBr),n(KBr))$ . solve simultaneous equations.
For eq. 3. divide both sides by 102.894 to isolate the moles of NaBr.
4. $1.1565\times n(KBr)+n(NaBr) = 5.4474\times 10^{-3}$
subtract eq. 2 to eliminate the $n(NaBr)$
$0.1565\times(n(KBr) = 0.2805\times 10^{-3} mol$ , so $nKbR = 1.792\times10^{-3} mol.$
So, $m(KBr) = 1.792\times10^{-3} mol.\times 119.002 g/mol=0.2133g$
So the mole fraction of KBr is $0.2133g/0.5605 g = 0.380$
the mass percent is $0.380 \times 100/100 = 38%$
What is the volume of a 0.563 M solution of NaOH that is required to react completely with 10.0 mL of a 0.255 M solution of HNO₃? [4.53 mL]
$10\;mL\times\frac{0.255\;mol/L}{0.563\;mol/L}=4.53 mL$
What volume of a 0.563 M solution of NaOH is required to react completely with 10.0 mL of a .255 M solution of acetic acid? [4.53 mL]
$10\;mL\times\frac{0.255\;mol/L}{0.563\;mol/L}=4.53 mL$
Note that this is the same answer as for the question above. It does not matter whether the acid is strong (HNO_3) or weak (CH_3COOH), a concept we will explore in later chapters.
What volume of a 0.563 M solution of NaOH is required to react completely with 10.0 mL of a .255 M solution of H₂SO₄? [9.06 mL]
$2\times 10\;mL\times\frac{0.255\;mol/L}{0.563\;mol/L}=9.06 mL$
this time, the difference occurs because the stoichiometry involving the diprotic acid $H_2SO_4$ is different from the other two monoprotic acids.

For even more fun, try these:

Review of Equations

Limiting Reagents

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