The combustion of methyl alcohol in an abundant excess of oxygen follows the equation:

2CH3OH + 3O2? 2CO2 + 4H2O

When 3.85 g of CH3OH was mixed with 6.15 g of O2 and ignited, 3.65 g of CO2 was obtained. What was the percentage yield of CO2?

The % yield of CO2 is 69.1 %

Explanation:

Step 1: Data given

Mass of CH3OH = 3.85 grams

Mass of O2 = 6.15 grams

Mass of CO2 obtained = 3.65 grams

Molar mass of CH3OH = 32.04 g/mol

Molar mass O2 = 32.00 g/mol

Molar mass CO2 = 44.01 g/mol

Step 2: The balanced equation

2CH3OH + 3O2→ 2CO2 + 4H2O

Step 3: Calculate moles  CH3OH

Moles CH3OH = mass CH3OH / molar mass CH3OH

Moles CH3OH = 3.85 grams / 32.04 g/mol

Moles CH3OH = 0.120 moles

Step 4: Calculate moles O2

Moles O2 = 6.15 grams / 32.00 g/mol

Moles O2 = 0.192 moles

Step 5: Calculate limtiting reactant

CH3OH is the limiting reactant. It will completely be consumed ( 0.120 moles).  O2 is in excess. There will react 3/2 * 0.120 = 0.180 moles

There will remain 0.192 - 0.180 = 0.012 moles

Step 6: Calculate moles CO2

For 2 moles CH3OH we need 3 moles O2 to produce 2 moles CO2 and 4 moles H2O

For 0.120 moles CH3OH we'll have 0.120 moles CO2

Step 7: Calculate mass CO2

Mass CO2 = moles CO2 * molar mass CO2

Mass CO2 = 0.120 moles * 44.01 g/mol

Mass CO2 = 5.28 grams

Step 8: Calculate % yield

% yield = (actual mass / theoretical mass) *100%

%yield = ( 3.65 grams / 5.28 grams ) *100%

% yield = 69.1 %

The % yield of CO2 is 69.1 %

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%