Respuesta :

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

Answer: The mass of [tex]MnO_2[/tex] required is 30.6 grams.

Explanation:

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

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]      .....(1)

Given mass of chlorine gas = 25.0 g

Molar mass of chlorine gas = 71 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of chlorine gas}=\frac{25.0g}{71g/mol}=0.352mol[/tex]

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

[tex]MnO_2+4HCl\rightarrow MnCl_2+Cl_2+2H_2O[/tex]

By Stoichiometry of the reaction:

1 mole of chlorine gas is produced when 1 mole of [tex]MnO_2[/tex] is reacted.

So, 0.352 moles of chlorine gas is produced when [tex]\frac{1}{1}\times 0.352=0.352mol[/tex] of  [tex]MnO_2[/tex] is reacted.

Now, calculating the mass of [tex]MnO_2[/tex] by using equation 1, we get:

Molar mass of [tex]MnO_2[/tex] = 87 g/mol

Moles of [tex]MnO_2[/tex] = 0.352 moles

Putting values in equation 1, we get:

[tex]0.352mol=\frac{\text{Mass of }MnO_2}{87g/mol}\\\\\text{Mass of }MnO_2=(0.352mol\times 87g/mol)=30.6g[/tex]

Hence, the mass of [tex]MnO_2[/tex] required is 30.6 grams.

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