Question

Question #5c3cd

Answer 1

To one significant figure, the mass percentage of Na₂CO₃ is 90 %.

The equations for the reaction are

Na₂CO₃•10H₂O → Na₂CO₃ + 10H₂O

Na₂CO₃ + 2HCl → 2NaCl + CO₂ + H₂O

Our conversions are:

Volume of HCl → moles of HCl → moles of Na₂CO₃ →
grams of Na₂CO₃ → mass % of Na₂CO₃

0.020 dm³ HCl × `(0.1 mol HCl)/(1 dm³ HCl)` = 2 × 10⁻³ mol HCl

Moles Na₂CO₃ = 2 × 10⁻³ mol HCl × `(1 mol Na₂CO₃)/(2 mol HCl)` = 1 × 10⁻³ mol Na₂CO₃

This gives the moles of Na₂CO₃ in 25 cm³ of prepared solution. In the original 250 cm³ of solution there must have been

1 × 10⁻³ mol Na₂CO₃ × `(250 cm³)/(25 cm³)` = 0.01 mol Na₂CO₃

Mass of Na₂CO₃ = 0.01 mol Na₂CO₃ × `(106.0 g Na₂CO₃)/(1 mol Na₂CO₃)` =
1 g Na₂CO₃

Mass % = (mass/total mass) × 100 % = `(1 g)/(1.2 g)` × 100 % = 90 %

Note: I have calculated the answer to only one significant figure, because that is all you gave for the concentration of the HCl. If you need more precision, you will have to recalculate.