Answer to Question #139591 in Chemistry for cyrrena rubio

Question

A silver nitrate solution contains 14.77 g of primary-standard AgNO3 in 1.00 L. What volume of this solution will be needed to react with 0.1957g of NaCl.

Answer 1

The balanced reaction equation is:

AgNO₃ + NaCl $\rightarrow$ NaNO₃ + AgCl$\downarrow$ .

As one can see, 1 mol of silver nitrate reacts with 1 mol of sodium chloride. The number of the moles of sodium chloride can be calculated from the ratio of its mass to its molar mass (58.44 g/mol):

$n(NaCl) = \frac{m(NaCl)}{M(NaCl)} = \frac{0.1957 \text{ g}}{58.44 \text{ g/mol}} = 0.003349$ mol.

Therefore, 0.003349 mol of silver nitrate will be needed to react with 0.1957g of NaCl. Let's calculate the mass of silver nitrate that contains 0.003349 mol (molar mass of AgNO₃ is 169.87 g/mol):

$m(AgNO_3) = n\cdot M = 0.003349\cdot169.87 = 0.5688$ g.

If a silver nitrate solution contains 14.77 g of AgNO₃ in 1L, then the volume needed is:

$V = \frac{0.5688 \text{ g }\cdot1 \text{L}}{14.77\text{ g}} = 0.03851$ L, or 38.5 mL.

Answer: 38.5 mL of the silver nitrate solution (14.77 g/L) will be needed to react with 0.1957 g of NaCl.