How would you calculate the % yield of hydrogen if 40.0 grams of magnesium react with an excess of nitric acid producing 1.70 grams of hydrogen gas. mg + 2hno3 → mg (no 3) 2 + h?

Question

Zoie Blanchard

Chemistry

20 December, 04:35

How would you calculate the % yield of hydrogen if 40.0 grams of magnesium react with an excess of nitric acid producing 1.70 grams of hydrogen gas. mg + 2hno3 → mg (no 3) 2 + h?

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Con · Answer 1

Magnesium in grams = 40 Actual yield is 1.70grams Calculating the yield based on the equation and moles, = 40g of Mn x (H2 (2.018g) / Mg (24.3g)) = 40 x 0.083 g H2 = 3.32g H2 Hydrogen % yield = (actual yield / calculated yield) x 100 = (1.70 / 3.32) x 100 Hydrogen % yield = 51.2%