Answer to Question #63151 in Chemistry for Glenn

Question

8) C6H14 + O2 à CO2 + H2O ​​​Mass of C6H14: 5.0 g
Experimental Mass of O2: 16.38g​​Experimental Yield of CO2: 14.96g​​Experiment Yield of H2O: 6.58g

9) Fe(OH)2 à FeO + H2O ​​​Mass of Fe(OH)2: 125.0g
Experimental Yield of FeO: 119.73g​​Experiment Yield of H2O: 28.93g

Answer 1

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