When 0.100 moles of bromine (Br_2) react with excess hydrogen gas (H_2) according to the equation Br_2 + H_2 → 2 HBr, how many grams of HBr will be produced?

• A. 3.65 g
• B. 18.1 g
• C. 36.2 g
• D. 8.05 g

Answer: (B)

Verified step by step guidance

1. Write down the balanced chemical equation: \(\mathrm{Br_2 + H_2 \rightarrow 2\ HBr}\).
2. Identify the given information: 0.100 moles of \(\mathrm{Br_2}\) and excess \(\mathrm{H_2}\), meaning \(\mathrm{Br_2}\) is the limiting reactant.
3. Use the mole ratio from the balanced equation to find moles of \(\mathrm{HBr}\) produced. According to the equation, 1 mole of \(\mathrm{Br_2}\) produces 2 moles of \(\mathrm{HBr}\), so multiply 0.100 moles of \(\mathrm{Br_2}\) by 2 to get moles of \(\mathrm{HBr}\).
4. Calculate the molar mass of \(\mathrm{HBr}\) by adding the atomic masses of hydrogen (approximately 1.01 g/mol) and bromine (approximately 79.90 g/mol).
5. Multiply the moles of \(\mathrm{HBr}\) by its molar mass to find the mass of \(\mathrm{HBr}\) produced in grams.