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%

We first assume that this gas is an ideal gas where it follows the ideal gas equation. the said equation is expressed as: pv = nrt. from this equation, we can predict the changes in the pressure, volume and temperature. if the volume and the temperature of this gas is doubled, then the pressure still stays the same.