A 5.000 g sample of a mixture which contains MgCO₃ and sand (SiO₂)was heated for 2.30 hours until no further reaction occurred. It was then cooled and the residue, which contained MgO and unchanged sand, was weighed. The reaction is:MgCO₃(s)? MgO(s)+ CO₂(g).The residue remaining (minus the CO₂, which was completely driven off)weighed 3.397 grams. Calculate the percent, by weight, of MgCO₃ in the original sample. Hint: Use moles as an intermediary when relating different units together.

Question

Quizplus · Accepted Answer

The weight loss is due to the decomposition of MgCO3 to MgO and CO2. The weight loss is 5.000 g - 3.397 g = 1.603 g, which corresponds to the weight of CO2 released. Using the molar masses of MgCO3 (84.31 g/mol) and CO2 (44.01 g/mol), calculate the moles of CO2 and then the moles of MgCO3. Convert moles of MgCO3 to grams and find the percentage in the original sample.

A 5000 G Sample of a Mixture Which Contains MgCO3 and and Sand

A 5000 G Sample of a Mixture Which Contains MgCO₃ and and Sand