Mixed Stoichiometry II Answer Key (CHEM 1405)

Back to Chemistry 1405 Practice Problems

Balance the following equations, and then solve the corresponding problems. Follow the rules for significant digits. Show all calculations.

1. 2C₄H₁₀ + 13O₂ 8CO₂ + 10H₂O

a) What mass of O₂ will react with 400.0 g C₄H₁₀?

$LaTeX: 400.0gC_4H_{10}\times(\frac{1molC_4H_{10}}{58.14gC_4H_{10}})\times(\frac{13molO_2}{2molC_4H_{10}})\times(\frac{32.00gO_2}{1molO_2})=$ $400.0gC_4H_{10}\times(\frac{1molC_4H_{10}}{58.14gC_4H_{10}})\times(\frac{13molO_2}{2molC_4H_{10}})\times(\frac{32.00gO_2}{1molO_2})=$ 1431g O₂

b) How many moles of water are formed in a)?

$LaTeX: 400.0gC_4H_{10}\times(\frac{1molC_4H_{10}}{58.14gC_4H_{10}})\times(\frac{10molH_2O}{2molC_4H_{10}})=$ $400.0gC_4H_{10}\times(\frac{1molC_4H_{10}}{58.14gC_4H_{10}})\times(\frac{10molH_2O}{2molC_4H_{10}})=$ 34.40g H₂O

2. 3HCl + Al(OH)₃ 3H₂O + AlCl₃

How many grams of aluminum hydroxide will react with 5.3 moles of HCl?

$LaTeX: 5.3molHCl\times(\frac{1molAl(OH)_2}{3molHCl})\times(\frac{78.01gAl(OH)_3}{1molAl(OH)_3})=$ $5.3molHCl\times(\frac{1molAl(OH)_2}{3molHCl})\times(\frac{78.01gAl(OH)_3}{1molAl(OH)_3})=$ 140g Al(OH)₃

3. Ca(ClO₃)₂ CaCl₂ + 3O₂

What mass of O₂ results from the decomposition of 1.00 kg of calcium chlorate?

$LaTeX: 1.00kgCa(ClO_3)_2\times(\frac{1000gCa(ClO_3)_2}{1kgCa(ClO_3)_2})\times(\frac{1molCa(ClO_3)_2}{206.98gCa(ClO_3)_2})\times(\frac{3molO_2}{1molCa(ClO_3)_2})\times(\frac{32.00gO_2}{1molO_2})=$ $1.00kgCa(ClO_3)_2\times(\frac{1000gCa(ClO_3)_2}{1kgCa(ClO_3)_2})\times(\frac{1molCa(ClO_3)_2}{206.98gCa(ClO_3)_2})\times(\frac{3molO_2}{1molCa(ClO_3)_2})\times(\frac{32.00gO_2}{1molO_2})=$ 464g O₂

4. The reaction of Ca with water can be predicted using the activity series. What mass of water is needed to completely react with 2.35 g of Ca?

Ca + 2H₂O Ca(OH)₂ + H₂

$LaTeX: 2.35gCa\times(\frac{1molCa}{40.00gCa})\times(\frac{2molH_2O}{1molCa})\times(\frac{18.02gH_2O}{1molH_2O})=$ $2.35gCa\times(\frac{1molCa}{40.00gCa})\times(\frac{2molH_2O}{1molCa})\times(\frac{18.02gH_2O}{1molH_2O})=$ 2.11g H₂O

5. Lead II nitrate reacts with potassium iodide to form lead II iodide and potassium nitrate.

a) Write the balanced equation for this reaction

Pb(NO₃)₂ + 2KI PbI₂ + 2KNO₃

b) How many moles of lead II iodide would be produced if 113 g of KI reacted with excess lead II nitrate?

$LaTeX: 113gKI\times(\frac{1molKI}{166.00gKI})\times(\frac{1molPbI_2}{2molKI})=$ $113gKI\times(\frac{1molKI}{166.00gKI})\times(\frac{1molPbI_2}{2molKI})=$ 0.340 mol PbI₂

c) How many grams of lead II nitrate would be needed to react with 113 g of KI?

$LaTeX: 113gKI\times(\frac{1molKI}{166.00gKI})\times(\frac{1molPb(NO_3)_2}{2molKI})\times(\frac{331.2gPb(NO_3)_2}{1molPb(NO_3)_2})=$ $113gKI\times(\frac{1molKI}{166.00gKI})\times(\frac{1molPb(NO_3)_2}{2molKI})\times(\frac{331.2gPb(NO_3)_2}{1molPb(NO_3)_2})=$ 113g Pb(NO₃)₂

6. Sulfuric acid (hydrogen sulfate) is made by reacting sulfur trioxide with water.

a) Write the balanced equation for the reaction stated above.

SO₃ + H₂O H₂SO₄

b) How many grams of sulfur trioxide would be needed to react with 4.23 mol H₂O?

$LaTeX: 4.23molH_2O\times(\frac{1molSO_3}{1molH_2O})\times(\frac{80.06gSO_3}{1molSO_3})=$ $4.23molH_2O\times(\frac{1molSO_3}{1molH_2O})\times(\frac{80.06gSO_3}{1molSO_3})=$ 339g SO₃

c) How many grams of sulfur trioxide would be needed to produce 2023 grams of sulfuric acid?

$LaTeX: 2023gH_2SO_4\times(\frac{1molH_2SO_4}{98.08gH_2SO_4})\times(\frac{1molSO_3}{1molH_2SO_4})\times(\frac{80.06gSO_3}{1molSO_3})=$ $2023gH_2SO_4\times(\frac{1molH_2SO_4}{98.08gH_2SO_4})\times(\frac{1molSO_3}{1molH_2SO_4})\times(\frac{80.06gSO_3}{1molSO_3})=$ 1651g H₂SO₄