Question

2AgNO3(aq)+CaCl2(aq)→2AgCl(s)+Ca(NO3)2(aq)
Part A
What mass of silver chloride can be produced from 1.20L of a 0.178M solution of silver nitrate?

Answer 1

Answer on Question #39835 - Chemistry - Other

Question

$$2 \mathrm{AgNO_3}(aq) + \mathrm{CaCl_2}(aq) \rightarrow 2 \mathrm{AgCl}(s) + \mathrm{Ca(NO_3)_2}(aq)$$

Part A

What mass of silver chloride can be produced from 1.20L of a 0.178M solution of silver nitrate?

Answer:

Number of moles of $\mathrm{AgNO_3}$ equals:

$$n = C \cdot V$$

C – Molar concentration of silver nitrate solution, $C = 0.178 \, \text{M}$.

V – Volume of a 0.178M solution silver nitrate, $V = 1.20 \, \text{L}$

$$n(\mathrm{AgNO_3}) = 0.178 \cdot 1.20 = 0.214 \, \text{moles}$$

Then we make a proportion:

- 2 moles of $\mathrm{AgNO_3}$ produce 2 moles of $\mathrm{AgCl}$

- 0.214 moles of $\mathrm{AgNO_3} - x$ moles of $\mathrm{AgCl}$

$$x = \frac{0.214 \cdot 2}{2} = 0.214 \, \text{moles}$$ $$n(\mathrm{AgCl}) = 0.214 \, \text{moles}$$

Molar mass of silver chloride equals:

$$M(\mathrm{AgCl}) = M(\mathrm{Ag}) + M(\mathrm{Cl}) = 108 + 35.5 = 143.5 \, \frac{\text{g}}{\text{mole}}$$

Then mass of silver chloride produced from 1.20L of a 0.178M solution of silver nitrate equals:

$$m(\mathrm{AgCl}) = n(\mathrm{AgCl}) \cdot M(\mathrm{AgCl}) = 0.214 \cdot 143.5 = 30.71 \, \text{g}$$

Answer: $m(\mathrm{AgCl}) = 30.71 \, \text{g}$