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Percent yield calculations differ from limiting stoichiometry problems by only one extra step. The question includes a mass recovered. Here is the formula you will use at the end of the problem. When you read the question the mass of the product must not be used until the very end. So.....put it in your pocket for later. You are allowed to use moles, grams, liters in this equation, as long as actual and theoretical are both in these units. EXAMPLE QUESTION #1 (Plain Stoichiometry not limiting) 2CO See how it says yields 48.3L of CO Now determine the volume of of CO Find the moles, the mol ratio, then find the volume of CO 69.1g CO(1 mol CO/28.0g)= 2.57mol CO 2.57mol CO(2mol CO 2.57mol CO Now take out that volume recovered and plug into the % yield equation. (48.3L/55.3L) x 100%= 77.3% yield Back to Math of Chemistry Links |