Question

What mass of Al is required to completely react with 25.0 g MnO2? What mass of is required to completely react with 25.0 ? 10.3 g Al 12.4 g Al 5.82 g Al 7.76 g Al

Answer 1

Answer: 10.3 g of Al is required to completely react with 25.0 g
`MnO_2`

Step-by-step explanation:

The balanced chemical equation for the reaction is:

`3MnO_2+4Al\rightarrow 3Mn+2Al_2O_3`

To find the moles we use the formula=
`\frac{\text{Given mass}}{\text {Molar mass}}`

moles of
`MnO_2=(25.0)/(87g/mol)=0.287moles`

According to stoichiometry:

3 moles of
`MnO_2` reacts with = 4 moles of aluminium

0.287 moles of
`MnO_2` reacts with =
`(4)/(3)* 0.287=0.383` moles of aluminium

Mass of aluminium =
`moles* {\text {Molar Mass}}=0.383mol* 27g/mol=10.3g`

Thus 10.3 g of Al is required to completely react with 25.0 g
`MnO_2`