Question

A carbonate sample with a mass of 5.095g is treated with 0.980g of hydrochloric acid. The resulting mixture of product and excess HCl has a mass of 5.435g. Calculate the % carbonate in the sample. Show your work.

Answer 1

Answer – Given, mass of sample =5.095 g , mass of HCl = 0.980 g

Mass of product + HCl = 5.435 g

As per the law of mass law of conservation of mass and it state that mass in the reaction neither created nor destroy, means

Mass of reacts side = mass of product side

So mass of reactant side = 5.095+0.980 = 6.075 g

Mass of product side = 5.435 g

We know the when carbonate reacted with HCl thee is evolved CO₂ gas from the carbonate group

So mas of CO₂ = 6.075 g – 5.435 g

                          = 0.64 g of CO₂

Moles of CO₂ = 0.64 g / 44 g.mol^-1

                       = 0.0145 moles

So, 1 moles of carbonate = 1 moles of CO₂

So, moles of carbonate = 0.0145 moles

Mass of carbonate = 0.0145 moles * 60.0 g/mol

                               = 0.872 g

So percent of carbonate = 0.872 g / 5.095 g *100 %

                                        = 17.1 %