CHEM 14A Lecture Notes - Lecture 6: Reagent
29 Nov 2015
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6. 00g of c react with 24 g of h2. Assume c = 12 and h = 1 for the sake of simplicity in this problem. Calculate the number of moles of c and h2 to find the limiting reagent. 24 g x 1 mole h2 = 12. From the balanced equation: 5 moles of c requires 0. 5 x 3/2 = 0. 75 moles of h2 for complete reaction. Since, 12 moles are present h2 is in excess and c is the limiting reagent. Therefore base calculation on moles of c present, namely 0. 5 moles. From balanced equation 0. 5 moles of c will form. 0. 5 x moles of c2h6 or more fully, 0. 5 moles of c x 1 mole c2h6 moles c. To find the volume of c2h6 use the fact that at. 25 c, a mole of any gas occupied 24 dm3.
