If law of conservation of mass holds good, 2.00 g of `Na_(2)SO_(4)` will react with 3.00 g of `BaCl_(2)` to produce `BaSO_(4)` equal to

Options

A. `5.00 g`
B. `3.36 g`
C. `3.29 g`
D. None of these

Answer

C

Explanation

In this problem, masses of both the reactants are given, hence it can be a problem of limiting quantities.
`2.00 g` of `Na_(2)SO_(4)=(2.00g)/(142gmol^(-1))=0.0141`mol
3.00 g of `BaCl_(2)=(3.00g)/(208,2gmol^(-1))0.0144`mol
According to the equation
`underset("1 mol")(BaCl_(2))+underset("1 mol")(Na_(2)SO_(4))tounderset("1 mol")(BaSO_(2))+2NaCl`
1 mol of `BaCl_(2)` reacts with 1 mol of `Na_(2)SO_(4)`
`:.` In this reaction, `BaCl_(2)` is in excess. In other words, `Na_(2)SO_(4)` is the limiting reactant.
`:.` Moles of `Na_(2)SO_(4)` produced
=Moles of `BaSO_(4)=0.0141`mol
`=(0.0141 mol)xx(233.4g mol^(-1))=3.29g`