Chemistry 10th Edition

Published by Brooks/Cole Publishing Co.
ISBN 10: 1133610668
ISBN 13: 978-1-13361-066-3

Chapter 3 - Chemical Equations and Reaction of Stoichiometry - Exercises - Calculations Based on Chemical Equations - Page 107: 20


(a) $ 8.00mol(O_2)$ (b) $6.40mol(NO)$ (c) $9.60mol(H_2O)$

Work Step by Step

1. Balance the equation: - Begin with the elements that appear only once in each side: Balance the $H$: The subscript for the hydrogen on the reactants is "3" from: $NH_3$. So, we should put a "$\frac{3}{2}$" as the coefficient of $H_2O$. * We divided by 2, because each water molecule has 2 hydrogens. $NH_3 + O_2 -- \gt NO + \frac{3}{2}H_2O$ Multiply all coefficients by 2: $2NH_3 + 2O_2 -- \gt 2NO + 3H_2O$ Balance the $O$: The products side has a total of 5 oxygens. Therefore we should put a "$\frac{5}{2}$" as the coefficient of $O_2$: $2NH_3 + \frac{5}{2}O_2 -- \gt 2NO + 3H_2O$ Multiply all coefficients by 2: $4NH_3 + 5O_2 -- \gt 4NO + 6H_2O$ ------- Use the balanced coefficients as conversion factors: (a) $6.40mol(NH_3) \times \frac{5(O_2)}{4(NH_3)} = 8.00mol(O_2)$ (b) $6.40mol(NH_3) \times \frac{4(NO)}{4(NH_3)} = 6.40mol(NO)$ (c) $6.40mol(NH_3) \times \frac{6(H_2O)}{4(NH_3)} = 9.60mol(H_2O)$
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