Question

Use the balanced equation to work the following problem:

CaCl2 + 2AgNO3 → 2 AgCl + Ca(NO3)2

How many grams of AgCl (molar mass=143 g/mol) is formed when 1.00 gram of AgNO3 (molar mass=170 g/mol) reacts?

1.00 g
0.00588 g
1.19 g
0.841 g

Answer 1

the answer is 0.841 g AgCl

Answer 2

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

Step-by-step 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}=(1.00g)/(170g/mol)=5.88* 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* 10^(-3)mol` of silver nitrate will produce =
`(2)/(2)* 5.88* 10^(-3)=5.88* 10^(-3)mol` of AgCl

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

Moles of AgCl =
`5.88* 10^(-3)mol`

Molar mass of AgCl = 143.32 g/mol

Putting values in equation 1, we get:

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

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