Question

For the reaction

Na2CO3+Ca(NO3)2⟶CaCO3+2NaNO3

how many grams of calcium carbonate, CaCO3, are produced from 19.9 g of sodium carbonate, Na2CO3?

Answer 1

Molar mass of Na2CO3,
MM = 2*MM(Na) + 1*MM(C) + 3*MM(O)
= 2*22.99 + 1*12.01 + 3*16.0
= 105.99 g/mol

mass of Na2CO3 = 19.9 g
mol of Na2CO3 = (mass)/(molar mass)
= 19.9/1.06*10^2
= 0.1878 mol


Balanced chemical equation is:
Na2CO3+Ca(NO3)2⟶CaCO3+2NaNO3


According to balanced equation
mol of CaCO3 formed = moles of Na2CO3
= 0.1878 mol


Molar mass of CaCO3,
MM = 1*MM(Ca) + 1*MM(C) + 3*MM(O)
= 1*40.08 + 1*12.01 + 3*16.0
= 100.09 g/mol


mass of CaCO3 = number of mol * molar mass
= 0.1878*1.001*10^2
= 18.79 g
Answer: 18.8 g