Answer:
The volume of CO2 is 1.53 L
Step-by-step explanation:
Step 1: Data given
Mass of formic acid = 3.15 grams
Volume of oxygen = 2.0 L
Molar mass of formic acid = 46.0 g/mol
Step 2: The balanced equation
2CH2O2 (aq) + O2 (g) → 2CO2 (g) + 2H2O (l)
Step 3: Calculate moles of formic acid
Moles formic acid = mass formic acid / molar mass formic acid
Moles formic acid = 3.15 grams / 46.0 g/mol
Moles formic acid = 0.0685 moles
Step 4: Calculate moles O2
22.4 L = 1.0 mol
2.0 L = 0.0893 moles
Step 5: Calculate limiting reactant
For 2 moles formic acid we need 1 mol O2 to produce 2 moles CO2 and 2 moles H2O
Formic acid is the limiting reactant. It will completely be consumed (0.0685 moles). O2 is in excess. There will react 0.0685 /2 = 0.03425 moles
There will remain 0.0893 - 0.03425 = 0.05505 moles O2
Step 6: Calculate moles CO2
For 2 moles formic acid we need 1 mol O2 to produce 2 moles CO2 and 2 moles H2O
For 0.0685 moles formic acid, we'll have 0.0685 moles CO2
Step 7: Calculate volume of CO2
1 mol = 22.4 L
0.0685 moles = 22.4 * 0.0685 = 1.53 L
The volume of CO2 is 1.53 L
NOTE: The balanced equation is 2HCHO2(aq) + O2(g) → 2CO2(g) + 2H2O(l)
not HCHO2(aq) + O2(g) → 2CO2(g) + 2H2O(l)