Question

Calculate the mass of MNO2 needed to produce 25.0 g of Cl2

Answer 1

Answer: The mass of
`MnO_2` required is 30.6 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}}` .....(1)

Given mass of chlorine gas = 25.0 g

Molar mass of chlorine gas = 71 g/mol

Putting values in equation 1, we get:

`\text{Moles of chlorine gas}=(25.0g)/(71g/mol)=0.352mol`

The chemical equation for the reaction of manganese (IV) oxide and hydrochloric acid follows:

`MnO_2+4HCl\rightarrow MnCl_2+Cl_2+2H_2O`

By Stoichiometry of the reaction:

1 mole of chlorine gas is produced when 1 mole of
`MnO_2` is reacted.

So, 0.352 moles of chlorine gas is produced when
`(1)/(1)* 0.352=0.352mol` of
`MnO_2` is reacted.

Now, calculating the mass of
`MnO_2` by using equation 1, we get:

Molar mass of
`MnO_2` = 87 g/mol

Moles of
`MnO_2` = 0.352 moles

Putting values in equation 1, we get:

`0.352mol=\frac{\text{Mass of }MnO_2}{87g/mol}\\\\\text{Mass of }MnO_2=(0.352mol* 87g/mol)=30.6g`

Hence, the mass of
`MnO_2` required is 30.6 grams.

Answer 2

I will solve this question assuming the reaction equation look like this:
MnO2 + 4 HCl ---> MnCl2 + Cl2 + 2 H2O.

For every one molecule of MnO2 used, there will be one molecule of Cl2 formed. If the molecular mass of MnO2 is 87g/mol and molecular mass of Cl2 is 73.0 g/mol, the mass of MnO2 needed would be:
Cl mass/Cl molecular mass * MnO2 molecular mass=
25g/ (73g/mol) * (87g/mol) * 1/1= 29.8 grams