Question

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

2C2H6 + 702 (arrow) 4CO2 + 6H20

1.5mol H20
1.6mol H20
1.0 mol h20
1.1 mol h20

Answer 1

`n_(H_2O)=1.5molH_2O`

Step-by-step 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*(1mol)/(30g) =0.50molC_2H_6\\n_(C_2H_6)^(reacted)=60.0gO_2*(1molO_2)/(32gO_2)*(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*(6molH_2O)/(2molC_2H_6)\\\\n_(H_2O)=1.5molH_2O`

Best regards.