Question

1. A mixture of 0.5 g CaCl2 and 0.5 g Na2CO3, was dissolved in water.
a. Which of the two reactants is limiting?
b. Calculate the mass of CaCO3 (s) that formed?

Answer 1

CaCl₂+Na₂CO₃=CaCO3+2NaCl

a. find the amount of substance

$n=\frac{m}{M}=\frac{0.5}{110.9840}=0.0045$

$n=\frac{m}{M}=\frac{0.5}{105.9884}=0.0047$

CaCl2 is limiting

b. CaCl2 is limiting, therefore the calculation is based on it

n(CaCL₂)=n(CaCO₃)=0.0045

$m=n\times M=0.0045\times100.0869=0.45$ g