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What volume of bromine gas (at STP) can be produced from 75.5L of chlorine gas

and 155 g of HBr?

1 Answer

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Answer: Your welcome!

Step-by-step explanation:

The volume of bromine gas (Br2) produced can be calculated as follows:

75.5 L of chlorine gas (Cl2) reacts with 155 g of HBr to produce Br2 and HCl.

Moles of Cl2 = 75.5 L × (1 mol/22.4 L) = 3.36 mol

Moles of HBr = 155 g/80.91 g/mol = 1.92 mol

The mole ratio of Cl2 to HBr is 3.36:1.92 = 1.75:1

The theoretical yield of Br2 is therefore 1.92 mol.

At STP, 1 mol of Br2 occupies a volume of 22.4 L.

Volume of Br2 produced = 22.4 L/mol × 1.92 mol = 43.2 L

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