Answered: Suppose 6.0 g of each reactant was used…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

Suppose 6.0 g of each reactant was used in the balanced reaction: 2Al (s) + 3Cl2 (g) = 2AlCl3 (s) Molar masses: Cl2=70.90 g/mol; AlCl3=133.3 g/mol; Al=26.98 g/mol Which is the limiting reagent? Which is the excess reagent? What is the theoretical yield? If 16.3 g of product was actually produced in an experiment, what is the percent yield?

Expert Solution

Step 1

Number of moles can be determined as follows:

$\begin{array}{rcl} \begin{matrix} No . of moles of Al = \frac{mass}{molarmass} \end{matrix} \\ = & \frac{6.0 g}{26.98 g / mol} \\ = & 0.22 mol \\ No . of moles of {Cl}_{2} & = & \frac{mass}{molarmass} \\ = & \frac{6.0 g}{70.90 g / mol} \\ = & 0.085 mol \end{array}$

Step 2

Limiting reagent and excess reagent can be calculated.

$2 Al + 3 {Cl}_{2} \to 2 {AlCl}_{3} \begin{matrix} 2 mol Al requires \to 3 mol {Cl}_{2} \\ 0.22 mol Al requires \to x mol {Cl}_{2} \\ No . of moles of {Cl}_{2} = \frac{0.22 molAl \times 3 mol {Cl}_{2}}{2 mol Al} \\ = 0.33 mol \\ \begin{matrix} No . of moles of {Cl}_{2} actually present = 0.085 mol \\ Limiting reagent is {Cl}_{2} \\ Excess reagent is Al \end{matrix} \end{matrix}$

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