Question

A sample of cesium carbonate, weighing 3.80 g, requires 1.90 g of hydrogen bromide gas to completely decompose to water, cesium bromide, and carbon dioxide gas. The total mass of water and cesium bromide formed is 5.20 g and no hydrogen bromide or cesium carbonate remains. According to the law of conservation of mass, what mass of carbon dioxide must have been formed?
A. 0.50 g
B 1.40 g
C 5.49 g
D 10.90 g
E 1.90 g

PLEASE GIVE ME AN EXPLANATION ON WHY IT IS THE CORRECT ANSWER.

Answer 1

Answer:

Choice A. 0.50 g.

Explanation:

According to the question, the reaction here converts

• caesium (cesium) carbonate and
• hydrogen bromide

to

• cesium bromide,
• carbon dioxide, and
• water.

By the Law of Conservation of Mass, matter can neither be created nor destroyed in a chemical reaction. (Shrestha et. al, Introductory & GOB Chemistry, Chemistry Libretexts, 2019.)

In other words, the mass of the reactants, combined, shall be the same as the mass of the products, combined.

What's the mass of the reactants?

$\rm \underbrace{\rm 3.80\;g}_{\mathrm{Cs_2CO_3}} + \underbrace{\rm 1.90\; g}_{\mathrm{HBr}} = 5.70\;g$.

What's the mass of the products?

Let $m(\mathrm{CO_2})$ represent the mass of carbon dioxide produced in this reaction.

The mass of the products will be:

$\rm \underbrace{\rm 5.20\;g}_{\mathrm{CeBr}\text{ and }\mathrm{H_2O}} + \underbrace{m(\mathrm{CO_2})}_{\mathrm{CO_2}}$.

The two masses shall be equal. That is:

$\rm 5.20\; g + \mathnormal{m}(\mathrm{CO_2}) = 5.70\;g$.

$m(\mathrm{CO_2}) = \rm 0.50\;g$.

In other words, by the Law of Conservation of Mass, the mass of carbon dioxide produced in this reaction will be $\rm 0.50\;g$.