Question

a 3.0g mixture of na2co3 and calcium trioxocarbonate(iv) liberated 0.44g of co2 on strong heating until no gas was evolved.what is the percentage of na2co3

Answer 1

Out of these two compounds only CaCO₃ decomposes upon heating according to the reaction

CaCO₃ --> CaO + CO₂

Therefore,

$m_{CaCO_3}=\frac{0.44g CO_2}{44gCO_2/mol}\times\frac{1molCaCO_3}{1molCO_2}\times\frac{100.1gCaCO_3}{1molCaCO_3}=1.0g$

$\%CaCO_3=\frac{1.0g}{3.0g}\times100\%=33\%$

$\%Na_2CO_3=100-33=67\%$

Answer: 67%