A 5.0 g sample of impure calcium carbonate ( CaCO ₃ ) containing 80 % pure CaCO ₃ by mass is allowed to react with 100 mL of 0.45 M H ₂ SO ₄ solution. What will be the mass of CO ₂ gas formed upon complete reaction? CaCO _ 3( s ) + H ₂ SO _ 4( aq ) CaSO _ 4( aq ) + H ₂ O _ ( l ) + CO _ 2( g )

Step-by-step solution

The balanced chemical equation is: CaCO _ 3( s ) + H ₂ SO _ 4( aq ) CaSO _ 4( aq ) + H ₂ O _ ( l ) + CO _ 2( g ) Moles of H ₂ SO ₄ available = Molarity Volume (in L) = 0.45 0.100 = 0.045 mol Mass of pure CaCO ₃ in the sample = 5.0 80 100 = 4.0 g Molar mass of CaCO ₃ = 100 g mol ⁻¹ Moles of pure CaCO ₃ available = 4.0 100 = 0.040 mol From the stoichiometry of the reaction, 1 mol of CaCO ₃ reacts with 1 mol of H ₂ SO ₄ . Since 0.040 mol of CaCO ₃ is less than 0.045 mol of H ₂ SO ₄ , CaCO ₃ is the limiting reagent. Mo