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sladkih [1.3K]
4 years ago
5

If the reaction of 9.87 moles of potassium with excess hydrobromic acid produced a 79.8% yield of hydrogen gas, what was the act

ual yield of hydrogen gas? unbalanced equation: k + hbr yields kbr + h2
Chemistry
1 answer:
Fed [463]4 years ago
7 0
Answer is: 3,94 of hydrogen gas.
Chemical reaction: 2K + 2HBr → 2KBr + H₂.
n(K) = 9,87 mol.
n(H₂) = ?.
from reaction: n(K) : n(H₂) = 2 : 1.
9,87 mol : n(H₂) = 2 : 1
n(H₂) = 4,935 mol for 100% yield of reaction
n(H₂) = 4,935 · 0,798 = 3,94 mol for 78,9 % yield of reaction.
n - amount of substance
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<h3>Answer:</h3>

2.125 g

<h3>Explanation:</h3>

We have;

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We are required to determine the mass of 9.51 g of a NaBr sample.

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In this case,

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A sample of 9.51 g of NaBr will also contain 22.345 of Na by mass

  • But;

% composition of an element by mass = (Mass of element ÷ mass of the compound) × 100

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Mass of sodium = (22.34% × 9.51 g) ÷ 100

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Thus, the mass of sodium in 9.51 g of NaBr is 2.125 g

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