Question

What volume of 0.305 m agno3 is required to react exactly with 155.0 ml of 0.274 m na2so4 solution? hint: you will want to write a balanced reaction?

Answer 1

The balanced chemical equation for reaction of $AgNO_{3}$ and $Na_{2}SO_{4}$ is as follows:

$2 AgNO_{3}+Na_{2}SO_{4}\rightarrow 2NaNO_{3}+Ag_{2}SO_{4}$

From the balanced chemical equation, 2 mol of $AgNO_{3}$ reacts with 1 mol of  $NaNO_{3}$.

First calculating number of moles of $NaNO_{3}$ as follows:

$M=\frac{n}{V}$

On rearranging,

$n=M\times V$

Here, M is molarity and V is volume. The molarity of $NaNO_{3}$  is given 0.274 M or mol/L and volume 155 mL, putting the values,

$n=0.274 mol/L\times 155\times 10^{-3}mL=0.04247 mol$

Since, 1 mol of $NaNO_{3}$  reacts with 2 mol of  $AgNO_{3}$ thus, number of moles of  $AgNO_{3}$  will be $2\times 0.04247 mol=0.08494 mol$.

Now, molarity of  $AgNO_{3}$ is given 0.305 M or mol/L thus, volume can be calculated as follows:

$V=\frac{n}{M}=\frac{0.08494 mol}{0.305 mol/L}=0.2785 L=278.5 mL$

Therefore, volume of  $AgNO_{3}$ is 278.5 mL.