Question

In an experiment, 3.25 g of C3H8 react with 3.50 g of O2.How many grams of H2O are formed?

Answer 1

1.57 g of H2O.

Step-by-step explanation:

What is given?

Mass of C3H8 = 3.25 g,

Mass of O2 = 3.50 g,

Molar mass of C3H8 = 44 g/mol,

Molar mass of O2 = 32 g/mol,

Molar mass of H2O = 18 g/mol.

Step-by-step solution:

First, let's see the chemical equation:

`C_3H_8+5O_2\rightarrow3CO_2+4H_2O.`

Let's convert the mass of each reactant to moles using their respective molar mass:

`\begin{gathered} 3.25\text{ g C}_3H_8\cdot\frac{1\text{ mol C}_3H_8}{44\text{ g C}_3H_8}=0.0739\text{ moles C}_3H_8, \\ \\ 3.50\text{ g O}_2\cdot\frac{1\text{ mol O}_2}{32\text{ g O}_2}=0.109\text{ moles O}_2. \end{gathered}`

Now let's see how many moles of H2O can be produced by each reactant. In the chemical equation 1 mol of C3H8 and 5 moles of O2 react to produce 4 moles of H2O:

`\begin{gathered} 0.0739\text{ moles C}_3H_8\cdot\frac{4\text{ moles H}_2O}{1\text{ mol C}_3H_8}=0.296\text{ moles H}_2O, \\ \\ 0.109\text{ moles O}_2\cdot\frac{4\text{ moles H}_2O}{5\text{ moles O}_2}=0.0872\text{ moles H}_2O. \end{gathered}`

You can note that the limiting reactant is O2. This is because O2 is being consumed first and this reactant imposes the limit to produce H2O.

The final step is to convert 0.0872 moles of H2O to grams using the molar mass of H2O, as follows:

`0.0872\text{ moles H}_2O\cdot\frac{1\text{8 g H}_2O}{1\text{ mol H}_2O}=1.57\text{ g H}_2O.`

The answer would be that 1.57 g of H2O are formed.