A 50 g mixture of CaCO₃ and SiO₂ was treated with excess HCl. If 8.96 L of CO₂ was released at STP, what is the mass fraction of SiO₂ in the mixture?
- 20%
- 80%
- 40%
- 10%
📝 Batafsil tushuntirish
Moles CO₂ = 8.96 / 22.4 = 0.4 mol.Moles CaCO₃ = 0.4 mol.
Mass CaCO₃ = 0.4 × 100 = 40 g.
Mass SiO₂ = 50 - 40 = 10 g.
Mass fraction = 10 / 50 = 0.2 or 20%.