Question

A chemist determined by measurements that 0.030 moles of barium participated in a chemical reaction. Calculate the mass of barium that participated in the chemical reaction. Round your answer to 2 significant digits.

Answer 1

Answer: 4.1 g of barium precipitated.

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar Mass}}$

Given : moles of barium = 0.030

Molar mass of barium = 137 g/mol

$0.030=\frac{x}{137}$

x= 4.1 g

Thus there are 4.1 g of barium that precipitated.