Mixed Stoichiometry I Answer Key (CHEM 1405)

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Mg      +      2HCl      Arrow.png      MgCl2      +      H2

Provide the mole ratio for each of the following mole conversions:

  1. Mg to H2:  LaTeX: (\frac{1molMg}{1molH_2})(1molMg1molH2) or LaTeX: (\frac{1molH_2}{1molMg})(1molH21molMg) 

  2. HCl to H2:   LaTeX: (\frac{2molHCl}{1molH_2})(2molHCl1molH2) or LaTeX: (\frac{1molH_2}{2molHCl})(1molH22molHCl) 

  3. Mg to HCl:   LaTeX: (\frac{1molMg}{2molHCl})(1molMg2molHCl) or LaTeX: (\frac{2molHCl}{1molMg})(2molHCl1molMg) 

  4. MgCl2 to HCl:   LaTeX: (\frac{1molMgCl_2}{2molHCl})(1molMgCl22molHCl) or LaTeX: (\frac{2molHCl}{1molMgCl_2})(2molHCl1molMgCl2) 

 

 

Ca3P2(s) + 6H2O(l) Arrow.png 3Ca(OH)2(s) + 2PH3(g)

  1. How many moles of phosphine (PH3) are prepared from 5.00 moles of Ca3P2 ?

 LaTeX: 5.00molCa_3P_2\times(\frac{2molPH_3}{1molCa_3P_2})=5.00molCa3P2×(2molPH31molCa3P2)= 10.0 mol PH3

 

  1. How many moles of phosphine (PH­3) are prepared from 5.00 moles of H2O?

 LaTeX: 5.00molH_2O\times(\frac{2molPH_3}{6molH_2O})=5.00molH2O×(2molPH36molH2O)= 1.67 mol PH3

 

 

C3H8      +      5O2      Arrow.png      3CO2      +      4H2O

  1. How many moles of CO2 are produced from the combustion of 45 g of C3H8?

LaTeX: 45gC_3H_8\times(\frac{1molC_3H_8}{44.11gC_3H_8})\times(\frac{3molCO_2}{1molC_3H_8})=45gC3H8×(1molC3H844.11gC3H8)×(3molCO21molC3H8)= 3.06 mol CO2

 

  1. What mass of H2O is produced if 0.200 mole of CO2 is also produced?

LaTeX: 0.200molCO_2\times(\frac{4molH_2O}{3molCO_2})\times(\frac{18.02gH_2O}{1molH_2O})=0.200molCO2×(4molH2O3molCO2)×(18.02gH2O1molH2O)= 4.81g H2O

 

  1. What mass of C3H8 is required to produce 1.80 g of H2O?

LaTeX: 1.80gH_2O\times(\frac{1molH_2O}{18.02gH_2O})\times(\frac{1molC_3H_8}{4molH_2O})\times(\frac{44.11gC_3H_8}{1molC_3H_8})=1.80gH2O×(1molH2O18.02gH2O)×(1molC3H84molH2O)×(44.11gC3H81molC3H8)= 1.10g C3H8

 

  1. What mass of C3H8 is required to react with 160 g of O2?

LaTeX: 160gO_2\times(\frac{1molO_2}{32.00gO_2})\times(\frac{1molC_3H_8}{5molO_2})\times(\frac{44.11gC_3H_8}{1molC_3H_8})=160gO2×(1molO232.00gO2)×(1molC3H85molO2)×(44.11gC3H81molC3H8)= 44g C3H8