Question

When 98.2 mL of 5.00 g/L AgNO3 is added to a coffee-cup calorimeter containing 98.2 mL of 5.00 g/L NaI, with both solutions at 25°C, what mass of AgI forms?

Answer 1

Answer: The mass of silver iodide formed is 0.681 grams

Explanation:

We are given:

Concentration of silver nitrate = 5.00 g/L

Concentration of sodium iodide = 5.00 g/L

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$     ......(1)

• For silver nitrate:

Volume of silver nitrate = 98.2 mL

Applying unitary method:

For 1000 mL of volume, the mass of silver nitrate is 5.00 grams

So, for 98.2 mL of volume, the mass of silver nitrate will be $\frac{5.00g}{1000mL}\times 98.2mL=0.491g$

Calculating the number of moles:

Given mass of silver nitrate = 0.491 g

Molar mass of silver nitrate = 169.9 g/mol

Putting values in equation 1, we get:

$\text{Moles of silver nitrate}=\frac{0.491g}{169.9g/mol}\\\\\text{Moles of silver nitrate}=0.0029mol$

• For NaI:

Volume of NaI = 98.2 mL

Applying unitary method:

For 1000 mL of volume, the mass of NaI is 5.00 grams

So, for 98.2 mL of volume, the mass of NaI will be $\frac{5.00g}{1000mL}\times 98.2mL=0.491g$

Calculating the number of moles:

Given mass of NaI = 0.491 g

Molar mass of NaI = 149.9 g/mol

Putting values in equation 1, we get:

$\text{Moles of NaI}=\frac{0.491g}{149.9g/mol}\\\\\text{Moles of silver nitrate}=0.0033mol$

The chemical equation for the reaction of silver nitrate and NaI follows:

$AgNO_3+NaI\rightarrow AgI+NaNO_3$

By Stoichiometry of the reaction:

1 mole of silver nitrate reacts with 1 mole of sodium iodide

So, 0.0029 moles of silver nitrate will react with = $\frac{1}{1}\times 0.0029=0.0029mol$ of sodium iodide

As, given amount of sodium iodide  is more than the required amount. So, it is considered as an excess reagent.

Thus, silver nitrate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of silver nitrate produces 1 mole of silver iodide

So, 0.0029 moles of silver nitrate will produce = $\frac{1}{1}\times 0.0029=0.0029moles$ of silver iodide.

Now, calculating the mass of silver iodide from equation 1, we get:

Molar mass of silver iodide = 234.8 g/mol

Moles of silver iodide = 0.0029 moles

Putting values in equation 1, we get:

$0.0029mol=\frac{\text{Mass of silver iodide}}{234.8g/mol}\\\\\text{Mass of silver iodide}=(0.0029mol\times 234.8g/mol)=0.681g$

Hence, the mass of silver iodide formed is 0.681 grams