If the reaction of 4.50 moles of sodium with excess hydrofluoric acid produced an 85.0% yield of hydrogen gas, what was the actual yield of hydrogen gas? Unbalanced equation: Na + HF "yields" / NaF + H2 1.91 mol H2 2.25 mol H2 2.65 mol H2 3.82 mol H2

Question

Chemistry

Paityn Castillo

5 September, 23:36

If the reaction of 4.50 moles of sodium with excess hydrofluoric acid produced an 85.0% yield of hydrogen gas, what was the actual yield of hydrogen gas? Unbalanced equation: Na + HF "yields" / NaF + H2 1.91 mol H2 2.25 mol H2 2.65 mol H2 3.82 mol H2

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Margaret Shields · Answer 1

First write a balanced chemical equation

that is 2Na + 2Hf - - - >2Naf + H2

This implies that to moles 2 of Na reacted with 2 moles of HF to form 2 moles of Naf and 1 mole of H2

thus the mole of H2 = 4.50 x1/2 = 2.25moles of H2

Paige · Answer 2

1.91 mol H2 Let's rewrite that equation into a balanced equation. Let's start with the unbalanced version: Na + HF = = > NaF + H2 We have equal amounts of Na and F on both sides, but twice as much H on the right. So let's double the Na, HF, and NaF coefficients. 2Na + 2HF = = > 2NaF + H2 And we now have equal quantities of all elements on both sides, so our equation is balanced. Now looking at the balanced equation, 2 moles of sodium should produce 1 mole of hydrogen gas. So we should have 4.50/2 = 2.25 moles of hydrogen gas. But we only have an 85% yield, so we multiply the expected yield by the percentage, giving 2.25 mol * 0.85 = 1.9125 mol. Rounding to 3 significant figures gives us 1.91 mol which matches "1.91 mol H2"