Answer:
a) CO is the limiting reactant.
b) [tex]n_{H_2}^{excess}=749.9592molH_2[/tex]
c) [tex]n_{CH_3OH}=0.0204molCH_3OH[/tex]
Explanation:
Hello,
In this case, the balanced chemical reaction:
[tex]CO + 2H_2 \rightarrow CH_3OH[/tex]
So we proceed as follows:
a) We first compute the moles of CO by using its density (0.00114 g/mL) and molar mass (28 g/mol):
[tex]n_{CO}=500mLCO*\frac{0.00114gCO}{1mLCO}*\frac{1molCO}{28gCO}=0.0204molCO[/tex]
Next, with the given 750 moles of hydrogen, we can compute the moles of carbon monoxide that are consumed by such amount of hydrogen by using their 1:2 molar ratio:
[tex]n_{CO}^{consumed\ by\ H_2}=750molH_2*\frac{1molCO}{2molH_2}=375molCO[/tex]
Thus, we see a clear excess of hydrogen, for that reason the carbon monoxide is the limiting reactant.
b) In this case, we first compute the moles of CO that are not consumed:
[tex]n_{CO}^{unchanged}=375mol-0.0204mol=374.9796molCO[/tex]
Next, we use the 1:2 molar ratio again to compute the unchanged moles of hydrogen which is the excess reactant:
[tex]n_{H_2}^{excess}=375.9796molCO*\frac{2molH_2}{1molCO} =749.9592molH_2[/tex]
c) Finally, we use the reacting moles of carbon monoxide to compute the formed moles of methanol by using the 1:1 molar ratio:
[tex]n_{CH_3OH}=0.0204molCO*\frac{1molCH_3OH}{1molCO} =0.0204molCH_3OH[/tex]
Best regards.
