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?
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%
