Question

Use the balanced equation to work the following problem:

Answer 1

The answer is 0.841 g AgCl

Answer 2

Answer: The mass of AgCl formed in the reaction is 0.841 grams.

Explanation:

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

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

Given mass of silver nitrate = 1.00 g

Molar mass of silver nitrate = 170 g/mol

Putting values in equation 1, we get:

$\text{Moles of silver nitrate}=\frac{1.00g}{170g/mol}=5.88\times 10^{-3}mol$

For the given chemical equation:

$CaCl_2+2AgNO_3\rightarrow 2AgCl+Ca(NO_3)_2$

By Stoichiometry of the reaction:

2 moles of silver nitrate produces 2 moles of AgCl

So, $5.88\times 10^{-3}mol$ of silver nitrate will produce = $\frac{2}{2}\times 5.88\times 10^{-3}=5.88\times 10^{-3}mol$ of AgCl

Now, calculating the mass of AgCl by using equation 1.

Moles of AgCl = $5.88\times 10^{-3}mol$

Molar mass of AgCl = 143.32 g/mol

Putting values in equation 1, we get:

$5.88\times 10^{-3}mol=\frac{\text{Mass of AgCl}}{143.32g/mol}\\\\\text{Mass of AgCl}=(5.88\times 10^{-3}mol\times 143.32g/mol)=0.841g$

Hence, the mass of AgCl formed in the reaction is 0.841 grams.