Question

Using the following equation: 3 Mg(OH)2 + 2 H3PO4 --> Mg3(PO4)2 + 6 H2OHow many grams of magnesium hydroxide do you need to produce 589 grams of water?

Answer 1

We have the following chemical equation:

`3Mg(OH)_2+2H_3PO_4\rightarrow Mg_3(PO_4)_2+6H_2O`

The first thing we do is chech if the equation is balanced. For this we count how many atoms of each element we have on each side of the equation:

Mg: 3

O: 14

H: 12

P: 2

So the equation is balanced.

Now, we transfor the mass of magnesium hydroxide and water into moles, by using their molar mass (M):

`M_(Mg(OH)_2)=M_(Mg)+2M_o+2M_H=24.3(g)/(mol)+2.16(g)/(mol)+2.1(g)/(mol)=58.3(g)/(mol)`
`M_(H_2O)=2M_H+M_O=2.1(g)/(mol)+16(g)/(mol)=18(g)/(mol)`

Now we calculate the number of moles of water (n) that we have to produce:

`n_(H_2O)=(589g)/(18(g)/(mol))=32.72mol`

Now, looking at the chemical equation we can see that every 3 moles of magnesium hydroxide that react produce 6 moles of water. So in order to produce 32.72 moles of water we need:

`n_(Mg(OH)_2)=(3molMg(OH)_2)/(6molH_2O).32.72molH_2O=16.36molMg(OH)_2`

Finally we transform the moles into grams:

`m_(Mg(OH)_2)=16.36mol.58.3(g)/(mol)=953.8g`

So the answer is 953.8g of magnesium hydroxide.