Answer to Question #63151 in Chemistry for Glenn
8) C6H14 + O2 à CO2 + H2O Mass of C6H14: 5.0 g
Experimental Mass of O2: 16.38gExperimental Yield of CO2: 14.96gExperiment Yield of H2O: 6.58g
9) Fe(OH)2 à FeO + H2O Mass of Fe(OH)2: 125.0g
Experimental Yield of FeO: 119.73gExperiment Yield of H2O: 28.93g
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2016-11-08T14:55:10-0500
8) Molar mass of C6H14 is 86.2 g/mol
Molar mass of O2 is 31.99 g/mol
Molar mass of CO2 is 43.99 g/mol
Molar mass of H2O is 17.99 g/mol
n = m/M
n(C6H14)=5.0/86.2=0.06 mol
n (O2) = 16.38/31.99 = 0.51 mol
n (CO2) = 14.96/43.99 = 0.34 mol
n (H2O) = 6.58/17.99 = 0.37 mol
According to this, let’s put coefficients to the equation:
2C6H14 + 19O2 12CO2 + 14H2O
8) Molar mass of Fe(OH)2 is 89.86 g/mol
Molar mass of FeO is 71.84 g/mol
Molar mass of H2O is 17.99 g/mol
n = m/M
n(Fe(OH)2)=125.0/89.86=1.39 mol
n (FeO) = 119.73/71.84 = 1.67 mol
n (H2O) = 28.93/17.99 = 1.61 mol
According to this, let’s put coefficients to the equation:
Fe(OH)2 FeO + H2O
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