Formic acid, HCHO2, burns in oxygen to form carbon dioxide and water as follows: HCHO2 (aq) + O2 (g) → 2 CO2 (g) + 2 H2O (l). If a 3.15-g sample of formic acid was burned in 2.0 L of oxygen, what volume of carbon dioxide would be produced? (Assume the reaction occurs at standard temperature and pressure, STP.)

The volume of CO2 is 1.53 L

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)

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the law of conservation of mass states that in a chemical reaction, the total mass of reactants is equal to the total mass of products. because atoms are only rearranged in a chemical reaction, there must be the same number of sodium atoms and chlorine atoms in both the reactants and products