Answer
6.9 g
Work Step by Step
1. There are two reactions in this process:
$$2Ba(s) + O_2(g) \longrightarrow 2 BaO(s)$$
$$BaO(aq) + H_2O(l) \longrightarrow Ba^{2+}(aq) + 2 OH^-(aq)$$
2. Calculate the amount of moles of $OH^-$:
$$1.0 \space L \times \frac{0.10 \space mol \space OH^-}{1 \space L} = 0.10 \space mol \space OH^-$$
3. Find the amount of moles of barium metal:
$$0.10 \space mol \space OH^- \times \frac{1 \space mol \space BaO}{2 \space mol \space OH^-} \times \frac{2 \space mol \space Ba}{2 \space mol \space BaO} = 0.050 \space mol \space Ba$$
$ Ba $ : 137.3 g/mol
- Using the molar mass as a conversion factor, find the mass in g:
$$ 0.050 \space mole \times \frac{ 137.3 \space g}{1 \space mole} = 6.9 \space g$$
