Question

If 15 g of C₂H₆ reacts with 60.0 g of O₂, how many moles of water (H₂O) will be produced?

Answer 1

Answer:

$n_{H_2O}=1.5molH_2O$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O$

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:

$n_{C_2H_6}^{available}=15g*\frac{1mol}{30g} =0.50molC_2H_6\\n_{C_2H_6}^{reacted}=60.0gO_2*\frac{1molO_2}{32gO_2}*\frac{2molC_2H_6}{7molO_2} =0.536molC_2H_6$

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:

$n_{H_2O}=0.50molC_2H_6*\frac{6molH_2O}{2molC_2H_6}\\\\n_{H_2O}=1.5molH_2O$

Best regards.