Question

what mass of Na₂CO₃ can be made if 1.52 g of NaHCO₃ is reacted.

2NaHCO₃(s)  →  Na₂CO₃(s) + H₂O(l) + CO₂(g)

Answer 1

According to the equation of the reaction

2NaHCO₃(s)  →  Na₂CO₃(s) + H₂O(l) + CO₂(g),

when 2 moles of NaHCO₃ react, 1 mole of Na₂CO₃ is formed:

$\frac{n(NaHCO_3)}{2} = n(Na_2CO_3)$ .

The number of the moles in 1.52 g of NaHCO₃ (M = 84.01 g/mol) is :

$n(NaHCO_3) = \frac{m}{M} = \frac{1.52 \text{ g}}{84.01\text{ g/mol}} = 0.0181$ mol.

Therefore, the number of the moles of Na₂CO₃ formed is:

$n(Na_2CO_3) = \frac{0.0181\text{ mol}}{2} = 0.0090$ mol.

Finally, the mass of Na₂CO₃ produced (M = 105.99 g/mol) is:

$m(Na_2CO_3) = n·M = 0.0090\text{ mol}·105.99\text{ g/mol}$

$m(Na_2CO_3) =0.9588$ , or 0.96 g.

Answer: 0.96 g of Na₂CO₃ can be made if 1.52 g of NaHCO₃ reacted.