Respuesta :
Answer:
We have the following chemical equation:
[tex]3Mg(OH)_2+2H_3PO_4\rightarrow Mg_3(PO_4)_2+6H_2O[/tex]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):
[tex]M_{Mg(OH)_2}=M_{Mg}+2M_o+2M_H=24.3\frac{g}{mol}+2.16\frac{g}{mol}+2.1\frac{g}{mol}=58.3\frac{g}{mol}[/tex][tex]M_{H_2O}=2M_H+M_O=2.1\frac{g}{mol}+16\frac{g}{mol}=18\frac{g}{mol}[/tex]Now we calculate the number of moles of water (n) that we have to produce:
[tex]n_{H_2O}=\frac{589g}{18\frac{g}{mol}}=32.72mol[/tex]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:
[tex]n_{Mg(OH)_2}=\frac{3molMg(OH)_2}{6molH_2O}.32.72molH_2O=16.36molMg(OH)_2[/tex]Finally we transform the moles into grams:
[tex]m_{Mg(OH)_2}=16.36mol.58.3\frac{g}{mol}=953.8g[/tex]So the answer is 953.8g of magnesium hydroxide.
