Question

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

Answer 1

`27"g"H_2O`

Explanation:

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

`C_2H_6"(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_2H_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_2O` 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`