Answer to Question #146489 in General Chemistry for Gusty

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?
2. What is the effect of heating the solution on the particle size of CaCO3 (s)?

Answer 1

A balanced equation for the reaction is

$CaCl_2+Na_2C\to CaCO_3+2NaCl$

1.

a.The limiting reagent is the reactant that has the least number of moles.

Number of moles of CaCl2

$Moles=\dfrac{0.5}{110.98}$

$= 0.00451$

Number of moles of Na2CO3

$Moles=\dfrac{0.5}{105.988}=0.00472$

Therefore, the limiting reagent in this case is $Na_2CO_3$

b. Calculate the mass of calcium carbonate using the balanced equation reaction ratio 1:1

Therefore number of moles of Calcium carbonate is 0.004.

$Mass =molar\ mass*number\ of\ moles$

$Mass =100.059*0.004$

Calcium carbonate mass = 0.4g

2.Upon heating the particle size of CaCO3 smaller particles favours higher crystalization and their sizes increases.