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).

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

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 $MgBr_2$

$\text{Mass of }MgBr_2=\text{Moles of }MgBr_2\times \text{Molar mass of }MgBr_2$

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

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