Question

For the following reaction, 5.61 grams of carbon monoxide are mixed with excess water . Assume that the percent yield of carbon dioxide is 85.3 %. carbon monoxide(g) water(l) carbon dioxide(g) hydrogen(g) What is the ideal yield of carbon dioxide

Answer 1

Answer: The ideal yield of carbon dioxide is 7.506 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$     .....(1)

Given mass of carbon monoxide = 5.61 g

Molar mass of carbon monoxide = 28 g/mol

Putting values in equation 1, we get:

$\text{Moles of carbon monoxide}=\frac{5.61g}{28g/mol}=0.200mol$

The chemical equation for the reaction of carbon monoxide and water follows:

$CO(g)+H_2O(l)\rightarrow CO_2(g)+H_2(g)$

By Stoichiometry of the reaction:

1 mole of carbon monoxide produces 1 mole of carbon dioxide

So, 0.200 moles of carbon monoxide will produce = $\frac{1}{1}\times 0.200=0.200mol$ of carbon dioxide

Now, calculating the mass of carbon dioxide from equation 1, we get:

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = 0.200 moles

Putting values in equation 1, we get:

$0.200mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=(0.200mol\times 44g/mol)=8.8g$

To calculate the experimental yield of carbon dioxide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Percentage yield of carbon dioxide = 85.3 %

Theoretical yield of carbon dioxide = 8.8 g

Putting values in above equation, we get:

$85.3=\frac{\text{Experimental yield of carbon dioxide}}{8.8g}\times 100\\\\\text{Experimental yield of carbon dioxide}=\frac{85.3\times 8.8}{100}=7.506g$

Hence, the ideal yield of carbon dioxide is 7.506 grams