Question

2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(l)

If 20.0 g of C2H6 react with 60.0 g of O2, assuming the reaction has gone to completion, calculate the mass (in grams) of water produced in this chemical reaction.

Answer 1

Let's calculate the amount of moles of reactants first:

$n​(C_2H_6)=\frac {m(C_2H_6)} {M(C_2H_6)}​=\frac {20 g} {30 g/mol} = 0.667 mol;$

$n​(O_2)=\frac {m(O_2)} {M(O_2)}​=\frac {60 g} {32 g/mol} = 1.875 mol;$

According to the stoichiometric quotients, we need 7/2=3.5 times more moles of oxygen comparing to ethane for complete reaction. In other words, we need to have 0.667 moles*3.5= 2.3345 moles of oxygen, while we have only 1.875 moles. This means, that ethane is in excess and we need to do our calculations according to the oxygen moles.

According to the stoichiometric quotients, the amount of water is 6/7 compared to moles of oxygen:

$n​(H_2O)=\frac {6} {7}n(O_2)​=\frac {6} {7} *1.875 mol= 1.607 mol;$

Now we can calculate the mass of water:

$m​(H_2O)=n(H_2O)​ * M(H_2O)=$

$=1.607 mol * 18 g/mol = 28.93 g;$

Answer: there will be 28.93g of water produced during this reaction.