If the reaction of 9.87 moles of potassium with excess hydrobromic acid produced a 79.8% yield of hydrogen gas, what was the actual yield of hydrogen gas? unbalanced equation: k + hbr kbr + h2 7.88 mol h2 5.74 mol h2 4.94 mol h2 3.94 mol h2

Question

Samantha Kaufman

Chemistry

30 October, 20:29

If the reaction of 9.87 moles of potassium with excess hydrobromic acid produced a 79.8% yield of hydrogen gas, what was the actual yield of hydrogen gas? unbalanced equation: k + hbr kbr + h2 7.88 mol h2 5.74 mol h2 4.94 mol h2 3.94 mol h2

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Answers (1)

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Audrey Sutton · Answer 1

Answer is: 3,94 mol of hydrogen gas.

Chemical reaction: 2K + 2HBr → 2KBr + H₂.

n (K) = 9,87 mol.

from reaction: n (K) : n (H₂) = 2 : 1.

9,87 mol : n (H₂) = 2 : 1.

n (H₂) = 4,935 mol at 100% of reaction.

n (H₂) = 4,395 mol · 0,798 = 3,938 mol.

n - amount of substance.