Question

37) The mole of AgCl formed, when I7g AgNO3 is mixed with 1.435 g NaCl is
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Answer 1

Answer: 0.25 moles

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\textMolar mass}}$  

For $AgNO_3$

Mass of  $AgNO_3$ given = 17 g

Molar mass of $AgNO_3$= g/mol

Putting values in above equation, we get:

Moles of $AgNO_3=\frac{17g}{170g/mol}=0.1mol$

For $NaCl$

Mass of  $NaCl$ given = 1.435 g

Molar mass of $NaCl$= 58.5 g/mol

Putting values in above equation, we get:

Moles of $NaCl=\frac{1.435g}{58.5g/mol}=0.25mol$

For the reaction of formation of sodium chloride, the equation follows:

$AgNO_3+NaCl\rightarrow AgCl+NaNO_3l$

By Stoichiometry of the reaction,

1 mole of $NaCl$ react with 1 mole of $AgNO_3$ to give 1 mole of $AgCl$

So, 0.25 $NaCl$ react with 0.25 mole of $AgNO_3$ to give 0.25 mole of $AgCl$

Thus $NaCl$ is the limiting reagent as it limits the formation of product.