Converting between mass and volume using moles | Exit Quiz | Oak National Academy

When converting the volume of a gas to the mass of a solid reactant in a balanced reaction, what must you first determine?

• the density of the gas

• the relative mass of the solid

• the number of moles of gas

• the temperature of the gas

Answer explanation

To use stoichiometry for conversion, you first need to calculate the number of moles of gas from its volume.

Order the following steps to calculate the mass of a solid reactant from the volume of a gas product in a chemical reaction. Starting with the volume of gas.

• Calculate the mass of the reactant using its relative mass.

• Calculate the number of moles of gas from its volume.

• Determine the volume of the gas.

• Use the stoichiometry of the balanced equation to find the moles of reactant.

Answer explanation

You first measure the volume of gas, convert it to moles, use the balanced equation to find the moles of the reactant, and finally calculate the mass of the reactant.

If 24 dm³ of hydrogen gas react with oxygen to form water, how many moles of oxygen gas were used? 2H₂ + O₂ → 2H₂O.

• 1 mole

• 2 moles

• 0.5 moles

• 1.5 moles

Answer explanation

At room temperature and pressure, one mole of any gas occupies 24 dm³. One mole of H₂ was used. For every one mole of hydrogen used, half the number of moles of oxygen are needed (2 : 1 ratio of H₂ : O₂).

A sample of oxygen gas weighing 16 grams is used in a reaction. Calculate the volume it occupies at room temperature and pressure. The RFM of oxygen (O₂) is 32. The molar gas volume is 24 dm³

• 12 dm³

• 24 dm³

• 36 dm³

• 384 dm³

• 0.67 dm³

Answer explanation

16 grams of oxygen gas occupy 12 dm³ at room temperature and pressure. 16 ÷ 32 = 0.5 moles (mass ÷ RFM = moles). 0.5 moles × 24 = 12 dm³.

Calculate the volume of CO₂ gas produced when 100 g of CaCO₃ decomposes. CaCO₃ → CaO + CO₂. Assume all the CaCO₃ reacts, the molar gas volume is 24 dm³. RFM of CaCO₃ = 100, CaO = 56, CO₂ = 44.

• 12 dm³

• 24 dm³

• 48 dm³

• 43 dm³

• 13 dm³

Answer explanation

First, calculate moles of CaCO₃ using mass ÷ RFM = moles. 100 g ÷ 100 = 1 mole. According to the symbol equation, 1 mole of CaCO₃ produces 1 mole of CO₂. At RTP 1 mol of a gas occupies 24 dm³. Volume = moles × 24; volume = 1 × 24 = 24 dm³.

Given 500 cm³ of oxygen gas at RTP, calculate the mass of H₂ gas that completely reacts with it to form water. Give answer to 2 s.f. 2H₂ + O₂ → 2H₂O. The molar gas volume is 24 dm³. RFM (Mr) of H₂ = 2

Answer explanation

Convert volume of O₂ to moles: 500 cm³ ÷ 24 000 cm³ = 0.02083 moles. For every 1 mole of O₂, 2 moles of H₂ are required. So 0.04166 moles of H₂ are needed. 0.04166 moles × 2 = 0.08332 g.