Answer:
[tex]n_{H_2O}=1.5molH_2O[/tex]
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
[tex]2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O[/tex]
Next, we identify the limiting reactant by computing the available moles of ethane and the moles of ethane consumed by 60.0 grams of oxygen:
[tex]n_{C_2H_6}^{available}=15g*\frac{1mol}{30g} =0.50molC_2H_6\\n_{C_2H_6}^{reacted}=60.0gO_2*\frac{1molO_2}{32gO_2}*\frac{2molC_2H_6}{7molO_2} =0.536molC_2H_6[/tex]
Thus, we notice there are less available moles, for that reason, the ethane is the limiting reactant. Finally, we can compute the produced moles of water by:
[tex]n_{H_2O}=0.50molC_2H_6*\frac{6molH_2O}{2molC_2H_6}\\\\n_{H_2O}=1.5molH_2O[/tex]
Best regards.
