Answer:
i. 3 moles.
ii. 132.03 grams.
iii. 105.6 grams.
Explanation:
Hello!
In this case, given the balanced chemical reaction, we can proceed as follows:
i. By starting with 5 moles of oxygen, via the 5:3 mole ratio we compute the produced moles of CO2:
[tex]n_{CO_2}=5molO_2*\frac{3molCO_2}{5molO_2} =3molCO_2[/tex]
ii. Now, since we have previously computed the moles of CO2 from the same moles of oxygen, by using its molar mass (44.02 g/mol), we obtain:
[tex]m_{CO_2}=3molCO_2*\frac{44.01gCO_2}{1molCO_2} =132.03gCO_2[/tex]
iii. Now, we need to combine the previously used two proportional factors for the calculation of the mass of CO2 from 128.00 grams of oxygen (molar mass 32.00 g/mol):
[tex]m_{CO_2}=128.00gO_2*\frac{1molO_2}{32.00gCO_2}*\frac{3molCO_2}{5molO_2}*\frac{44.01gCO_2}{1molCO_2}\\\\m_{CO_2}=105.6gCO_2[/tex]
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