Find the attached file For the following reaction, 4.59 grams of water are mixed with excess carbon monoxide. The reaction yields 8.28 grams of carbon dioxide.carbon monoxide (g) + water (1) >carbon dioxide (g) + hydrogen (g)What is the theoretical yield of carbon dioxidegramsWhat is the percent vield of carbon dioxide ?

Answered: For the following reaction, 4.59 grams…

World of Chemistry, 3rd edition

3rd Edition

ISBN:9781133109655

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

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

Chapter9: Chemical Quantities

Section: Chapter Questions

Problem 52A

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Find the attached file

For the following reaction, 4.59 grams of water are mixed with excess carbon monoxide. The reaction yields 8.28 grams of carbon dioxide.
carbon monoxide (g) + water (1) >carbon dioxide (g) + hydrogen (g)
What is the theoretical yield of carbon dioxide
grams
What is the percent vield of carbon dioxide ?

Expert Solution

Step 1

Percentile yield is defined as the ratio of actual yield by theoretical yield multiplied by 100. Actual yield is the mass of the product produced in the lab while the theoretical yield is the mass of the product produced according to the mass of the reactant.

Step 2

The formula to calculate the moles of water is as follows:

$M o l e s o f w a t e r = \frac{M a s s}{M o l a r M a s s}$

Substitute $4.59 g$ for mass and $18.01528 g / m o l$ for molar mass to calculate the moles of water.

$\begin{array}{rcl} M o l e s o f w a t e r & = & \frac{4.59 g}{18.01528 g / m o l} \\ = & 0.254783717 m o l \end{array}$

Carbon monoxide is present in excess and therefore water acts as the limiting reagent and therefore the amount of $C O_{2}$ produced depends on the amount of limiting reagent water.

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