Question

in a single displacement reaction of zinc and silver nitrate, how many moles of zinc are required in this reaction when 4 g of silver nitrate is present?

Answer 1

Answer: The amount of zinc required are 0.0118 moles

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 = 4 g

Molar mass of silver nitrate = 169.9 g/mol

Putting values in above equation, we get:

`\text{Moles of silver nitrate}=(4g)/(169.9g/mol)=0.0235mol`

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

`Zn+2AgNO_3\rightarrow 2Ag+Zn(NO_3)_2`

By Stoichiometry of the reaction:

2 moles of silver nitrate reacts with 1 mole of zinc

So, 0.0235 moles of silver nitrate will react with =
`(1)/(2)* 0.0235=0.0118mol` of zinc

Hence, the amount of zinc required are 0.0118 moles