Question

How many grams of MgBr2 are needed to produce 75g or metal?

Answer 1

Answer : The mass of
`MgBr_2` needed to produce are, 1150 grams

Solution : Given,

Mass of metal (magnesium) = 75 g

Molar mass of magnesium = 24 g/mole

Molar mass of
`MgBr_2` = 184 g/mole

First we have to calculate the moles of metal (magnesium).

`\text{Moles of Mg}=\frac{\text{Mass of Mg}}{\text{Molar mass of Mg}}=(75g)/(24g/mole)=6.25moles`

Now we have to calculate the moles of
`MgBr_2`.

The balanced chemical reaction will be,

`Mg+Br_2\rightarrow MgBr_2`

From the balanced reaction we conclude that

1 moles of magnesium react to give 1 mole of
`MgBr_2`

6.25 moles of magnesium react to give 6.25 moles of
`MgBr_2`

Now we have to calculate the mass of
`MgBr_2`

`\text{Mass of }MgBr_2=\text{Moles of }MgBr_2* \text{Molar mass of }MgBr_2`

`\text{Mass of }MgBr_2=(6.25mole)* (184g/mole)=1150g`

Therefore, the mass of
`MgBr_2` needed to produce are, 1150 grams