Chemistry 10th Edition

by Whitten, Kenneth W.; Davis, Raymond E.; Peck, Larry; Stanley, George G.

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 - Conceptual Exercises - Page 112: 96

Answer

$V(HNO_3) = 68ml$

Work Step by Step

Considering stoichiometric coefficients, we can conclude the following: $m(HNO_3) = \frac{m(Bi)}{A(Bi)}\cdot 4\cdot M(HNO_3) = \frac{20g}{209\frac{g}{mol}}\cdot 4\cdot 63\frac{g}{mol}=24.11g$ $m(HNO_3) = \rho V\omega \implies V=\frac{m(HNO_3)}{\rho \omega} = \frac{24.11g}{1.182\frac{g}{ml}\cdot 0.3} = 68ml$

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