In the reaction Na2CO3 + 2HCl → 2NaCl + CO2 + H2O, how many grams of CO2 are produced when 7.5 moles of HCl is fully reacted?

165 of CO₂.

Explanation:

In the reaction:

Na2CO3 + 2HCl → 2NaCl + CO2 + H2O

2 moles of HCl reacts producing 1 mole o CO₂

If 7.5 moles of HCl reacts, moles of CO₂ produced are:

7.5 moles of HCl ₓ ( 1 mol CO₂ / 2 mol HCl) = 3.75 mol CO₂. As molar mass of CO₂ is 44g/mol, mass of CO₂ is:

3.75 mol CO₂ ₓ (44g / 1mol) = 165 of CO₂

Explanation:

Equation of the reaction

Na2CO3 + 2HCl → 2NaCl + CO2 + H2O

Given:

Number of moles of HCl, nA = 7.5 moles

By stoichiometry, 2 moles of HCl reacted fully to liberate 1 mole of CO2 gas.

Therefore, number of moles of CO2, ng = 7.5/2 × 1 moles

= 3.75 moles

Mass = molar mass × number of moles

Molar mass of CO2 = 12 + (16 × 2)

= 44 g/mol

Mass of CO2 = 44 × 3.75

= 165 g of CO2

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