To calculate the volume of oxygen produced when 10 g of silver nitrate (AgNO₃) is decomposed, we first need to determine the balanced chemical reaction for the decomposition of silver nitrate. The decomposition of silver nitrate can be represented as follows:
\[ 2 \text{AgNO}_3 (s) \rightarrow 2 \text{Ag} (s) + 2 \text{NO}_2 (g) + \text{O}_2 (g) \]
From the balanced equation, we see that 2 moles of silver nitrate produce 1 mole of oxygen gas.
Next, we'll calculate the molar mass of silver nitrate (\( \text{AgNO}_3 \)):
\[ \text{Molar mass of AgNO}_3 = \text{Ag} + \text{N} + 3 \times \text{O} = 108 + 14 + 3 \times 16 = 108 + 14 + 48 = 170 , \text{g/mol} \]
Now we will calculate the number of moles of silver nitrate in 10 g:
\[ \text{Number of moles of AgNO}_3 = \frac{10 , \text{g}}{170 , \text{g/mol}} = \frac{10}{170} \approx 0.0588 , \text{mol} \]
According to the stoichiometry of the reaction, 2 moles of AgNO₃ produce 1 mole of O₂ gas. Therefore, the number of moles of O₂ produced from 0.0588 moles of AgNO₃ can be calculated as follows:
\[ \text{Number of moles of O}_2 = \frac{0.0588 , \text{mol AgNO}_3}{2} \approx 0.0294 , \text{mol O}_2 \]
Now, we can calculate the volume of oxygen gas produced at standard temperature and pressure (s.t.p), where 1 mole of gas occupies 22.4 dm³:
\[ \text{Volume of O}_2 = \text{Number of moles of O}_2 \times 22.4 , \text{dm}^3/\text{mol} = 0.0294 , \text{mol} \times 22.4 , \text{dm}^3/\text{mol} \]
Calculating this gives:
\[ \text{Volume of O}_2 \approx 0.65856 , \text{dm}^3 \]
Rounding to three significant figures, the volume of oxygen produced is approximately:
\[ \boxed{0.659 , \text{dm}^3} \]