Question

What is the mass of silver chlorate (191.32 g/mol) that releases 0.466 L of oxygen gas at STP

Answer 1

What is the mass of silver chlorate (191.32 g/mol) that releases 0.466 L of oxygen gas at STP.

**Solution.**

Write the chemical reaction path:

$$2AgClO_3 \rightarrow 2AgCl + 3O_2\uparrow;$$

2 moles of silver chlorate $(AgClO_3)$ releases 3 moles of oxygen. The volume of 1 mole of gas at STP is 22.41 L. So find that:

$$2 * 191.32\ \mathrm{g}\left(AgClO_3\right) - 3 * 22.41\ \mathrm{L}\left(O_2\right)$$

And $x\ g$ of silver chlorate releases 0.466 L of oxygen

$$x\ \mathrm{g}\left(AgClO_3\right) - 0.446\ \mathrm{L}\left(O_2\right);$$

Find $x$:

$$x = \frac{2 * 191.32 * 0.446}{3 * 22.4} = 2.54\ \mathrm{g}\left(AgClO_3\right);$$

Answer: $2.54\ \mathrm{g}\left(AgClO_3\right)$