Question

what mass of sodium chloride will be required to completely react with 4.00×10²⁴ formula units of agno₃ and form agcl and sodium nitrate?
0.114 g
8.80 g
6.64 g
388 g

Explanation:

Step1: Write the balanced chemical equation

$NaCl + AgNO_3
ightarrow AgCl+NaNO_3$
The mole - ratio of $NaCl$ to $AgNO_3$ is 1:1.

Step2: Calculate the number of moles of $AgNO_3$

We know that 1 mole of any substance contains $6.022\times 10^{23}$ formula units.
Let $n$ be the number of moles of $AgNO_3$.
$n=\frac{4.00\times 10^{24}\text{ formula units}}{6.022\times 10^{23}\text{ formula units/mol}}\approx6.64\text{ mol}$

Step3: Determine the number of moles of $NaCl$

Since the mole - ratio of $NaCl$ to $AgNO_3$ is 1:1, the number of moles of $NaCl$ is also $n_{NaCl}=6.64\text{ mol}$

Step4: Calculate the mass of $NaCl$

The molar mass of $NaCl$ is $M = 22.99\text{ g/mol}+35.45\text{ g/mol}=58.44\text{ g/mol}$
The mass of $NaCl$, $m = n\times M$
$m=6.64\text{ mol}\times58.44\text{ g/mol}\approx388\text{ g}$

Answer:

388 g