Calculate the number of moles of calcium chloride that can be obtained from 50 g of limestone in the presence of excess hydrogen chloride.
Calculate the number of moles of calcium chloride that can be obtained from 50 g….
CaCO₃(s) + 2 HCl(aq) —> CaCl₂(aq) + CO₂(g) + H₂O(l)
From the balanced equation, 1 mole of calcium carbonate (CaCO₃) reacts with 2 moles of HCl to produce 1 mole of calcium chloride (CaCl₂).
Molar masses:
CaCO₃ = Ca(40.08) + C(12.01) + 3O(16.00) = 100.09 g/mol
CaCl₂ = Ca(40.08) + 2Cl(35.45) = 110.98 g/mol
Now, let’s calculate the number of moles of CaCO₃ in 50g:
Number of moles of CaCO₃ = Mass / Molar mass
Number of moles of CaCO₃ = 50g / 100.09 g/mol ≈ 0.4998 mol
Since the mole ratio of CaCO₃ to CaCl₂ is 1:1, the number of moles of CaCl₂ that can be obtained is also approximately 0.4998 mol.
Thus, about 0.4998 moles of calcium chloride can be obtained from 50g of limestone in the presence of excess hydrogen chloride.
