Question

Stoichiometry!

Please note:
- Use 6.022x1023 for avogadro’s number
- Ignore sig figs and do not round the final answer.
- Keep it to 1 decimal place.

Answer 1

Answer:

a) 13.2 moles $2H_{2}O$

b) 79.33 grams of $2H_{2}O$

Explanation:

First, we'll need to balance the equation

$H_{2(g)} + O_{2(g)}$ → $H_{2}O_{(g)}$

There are 2 (O) on the left and only one on the right, so we'll add a 2 coefficient to the right.

$H_{2(g)} + O_{2(g)}$ → $2H_{2}O_{(g)}$

Now there are 4 (H) on the right and only 2 on the left, so we'll add a 2 coefficient to the ($H_{2}$) on the left.

$2H_{2(g)} + O_{2(g)}$ → $2H_{2}O_{(g)}$

The equation is now balanced.

a) This can be solved with a simple mole ratio.

4.6 moles $O_{2}$ × $\frac{2 moles H_{2}O}{1 mole O_{2}}$ = 13.2 moles $2H_{2}O$

b) This problem is solved the same way!

2.2 moles $H_{2}$ × $\frac{2 moles H_{2}O}{2 moles H_{2}}$ = 2.2 moles $2H_{2}O$

However, this problem wants the mass of $2H_{2}O$, not the moles.

The molecular weight of $2H_{2}O$ is the weight of 4 (H) molecules and 2 (O) molecules (found on the periodic table). So,

4(1.008) + 2(15.999) = 36.03 g/mol

2.2 moles $2H_{2}O$ × $\frac{36.03 g}{1 mol}$ = 79.33 grams of $2H_{2}O$