Calculate the volume of oxygen produced when 10g of silver nitrate was completely decomposed by heating at s.t.p(Ag=108,N=14,O=16)Molar gas volume at s.t.p is 22.4dm3

Answer 1

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:

$$2{\text{AgNO}}_3(s)\to 2\text{Ag}(s)+2{\text{NO}}_2(g)+{\text{O}}_2(g)$$

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:

$$\boxed{{\displaystyle 0.659,{\text{dm}}^3}}$$