Question

2Na+2H2O➡️2NaOH+H2In an experiment 2.35 g of Na react with 7.25 g of H2O.How many grams of H2 are formed?

Answer 1

0.102 g of H2.

Step-by-step explanation:

What is given?

Mass of Na = 2.35 g,

Mass of H2O = 7.25 g,

Molar mass of Na = 23 g/mol,

Molar mass of H2O = 18 g/mol,

Molar mass of H2 = 2 g/mol.

Step-by-step solution:

First, let's convert each mass of reactant to moles using their respective molar mass, like this:

`\begin{gathered} 2.35\text{ g Na}\cdot\frac{1\text{ mol Na}}{23\text{ g Na}}=0.102\text{ moles Na,} \\ \\ 7.25\text{ g H}_2O\cdot\frac{1\text{ mol H}_2O}{18\text{ g H}_2O}=0.403\text{ moles H}_2O. \end{gathered}`

Now, let's see how many moles of H2 can be produced by each number of moles of each reactant.

You can see that 2 moles of Na reacted produces 1 mol of H2, and 2 moles of H2O reacted produces 1 mol of H2:

`\begin{gathered} 0.102\text{ moles Na}\cdot\frac{1\text{ mol H}_2}{2\text{ moles Na}}=0.051\text{ moles H}_2, \\ \\ 0.403\text{ moles H}_2O\cdot\frac{1\text{ mol H}_2}{2\text{ moles H}_2O}=0.202\text{ moles H}_2. \end{gathered}`

You can realize that the limiting reactant, in this case, is Na because this reactant is consumed first and imposes the limit to produce H2.

The final step is to convert 0.051 moles of H2 to grams using the molar mass of H2, as follows:

`0.051\text{ moles H}_2\cdot\frac{2\text{ g H}_2}{1\text{ mol H}_2}=0.102\text{ g H}_2.`

The answer would be that 0.102 g of H2 are formed.