# If 15 g of C_2H_6 react with 45 g of O_2, how many grams of water will be produced?

May 22, 2017

$27 \text{g} {H}_{2} O$

#### Explanation:

Let's write the chemical equation for this combustion reaction:

${C}_{2} {H}_{6} \text{(g)" + 7/2O_2"(g)" rarr 2CO_2"(g)" + 3H_2O"(g)}$

Since we're given the amounts of more than one reactant, we must determine which one is limiting by using their molar masses to convert to moles, and then dividing that number by the coefficient in the equation:

15cancel("g"C_2H_6)((1"mol"C_2H_6)/(30.08cancel("g"C_2H_6))) = 0.50"mol"C_2H_6

75cancel("g"O_2)((1"mol"O_2)/(32.00cancel("g" O_2))) = (2.344"mol")/(7/2"(coefficient)") = 0.67"mol"O_2

Since ${C}_{2} {H}_{6}$ is present in relative deficiency, it is the limiting reactant.

Now, we'll use the stoichiometric relationships (the coefficients) to find the moles of ${H}_{2} O$ and then water's molar mass to calculate the mass formed.

0.50cancel("mol"C_2H_6)((3cancel("mol"H_2O))/(1cancel("mol"C_2H_6))((18.02"g"H_2O)/(1cancel("mol"H_2O))) = color(red)(27"g"H_2O