Question

What is the percent yield if 15.5 g SO2 is obtained from the reaction of 42.5 g of O2 with excess ZNS according to the following equation 2ZnS (s) + 3O2 -> 2ZnO (s) + 2 SO2(g)

Answer 1

Percent yield = 27.3%.

Step-by-step explanation:

First, let's write the chemical equation:

`2ZnS+3O_2\rightarrow2ZnO+2SO_2.`

The limiting reactant, in this case, would be O2 because we have an excess of ZnS. So, we have to convert 42.5 g of O2 to moles. Remember that the molar mass of O2 is 32 g/mol (you can calculate the molar mass of a compound using the periodic table). The conversion will be:

`42.5\text{ g O}_2\cdot\frac{1\text{ mol O}_2}{32\text{ g O}_2}=1.33\text{ moles O}_2.`

With this value, we're going to find the number of moles of SO2 produced by 1.33 moles of O2. You can see in the chemical equation that 3 moles of O2 reacted produces 2 moles of SO2, so the calculation would look like this:

`1.33\text{ moles O}_2\cdot\frac{2\text{ moles SO}_2}{3\text{ moles O}_2}=0.887\text{ moles SO}_2.`

The next step is to find the mass of SO2 based on its number of moles and the molar mass of SO2 which is 64 g/mol, like this:

`0.887\text{ moles SO}_2\cdot\frac{64\text{ g SO}_2}{1\text{ mol SO}_2}=56.8\text{ g SO}_2.`

And finally, we replace the values that we have in the formula of percent yield:

`\%\text{ yield }=\frac{experimental\text{ yield}}{theoretic\text{al yield}}\cdot100\%.`

Our experimental yield is the mass that we obtained of SO2 which is 15.5 g and the theoretical yield is the mass that we found through stoichiometry which is 56.8g:

`\%\text{ yield}=\frac{15.5\text{ g}}{56.8\text{ g}}\cdot100\%\approx27.3\%.`

The percent yield of this reaction would be 27.3%