Question

What volume of Br2 gas (at STP) is created when 100.0 g of chromium (III) bromide decomposes according to the reaction?

2CrBr3-> 2CrBr2 + Br2

Answer 1

4.2L

Step-by-step explanation:

We can use stoichiometry and the ideal gas law to determine the volume of Br2 gas produced at STP.

First, we need to determine the number of moles of CrBr3 used in the reaction. We can do this by dividing the given mass by the molar mass of CrBr3:

n(CrBr3) = 100.0 g / 266.81 g/mol = 0.375 mol

According to the balanced chemical equation, 2 moles of CrBr3 produces 1 mole of Br2 gas. Therefore, the number of moles of Br2 gas produced in the reaction is:

n(Br2) = 0.375 mol / 2 = 0.1875 mol

At STP (standard temperature and pressure), one mole of an ideal gas occupies 22.4 L. Therefore, the volume of Br2 gas produced at STP is:

V(Br2) = n(Br2) * 22.4 L/mol = 0.1875 mol * 22.4 L/mol = 4.2 L

Therefore, the volume of Br2 gas produced at STP when 100.0 g of CrBr3 decomposes is 4.2 L.