Mg3N2(s)+6H2O(l) to 3Mg(OH)2(s)+2NH2(g)how many grams of H2O are needed to producce 150g of Mg(OH)2?
you need twice the number of moles of water than you get of Magnsium hydroxide.
calculate the moles of 150g magnesium hydroxide, double it, tha tis the moles of water, then mulitpy by 18 to get grams of water.
17g
Confirmed
To determine the amount of water (H2O) needed to produce a certain amount of magnesium hydroxide (Mg(OH)2), we need to use stoichiometry. Here's how you can calculate it:
1. Write and balance the chemical equation:
Mg3N2(s) + 6H2O(l) → 3Mg(OH)2(s) + 2NH3(g)
2. Determine the molar mass of Mg(OH)2:
Mg: 24.31 g/mol
O: 16.00 g/mol (x 2)
H: 1.01 g/mol (x 2)
Molar mass of Mg(OH)2 = 24.31 + (16.00 x 2) + (1.01 x 2) = 58.33 g/mol
3. Convert the given mass of Mg(OH)2 to moles:
Moles of Mg(OH)2 = Mass / Molar mass
Moles of Mg(OH)2 = 150 g / 58.33 g/mol ≈ 2.57 mol
4. Now, let's use the stoichiometry from the balanced equation to find the moles of H2O:
From the equation, we know that:
3 moles of Mg(OH)2 react with 6 moles of H2O
So, 1 mole of Mg(OH)2 reacts with (6/3) moles of H2O
Moles of H2O = Moles of Mg(OH)2 x (6/3)
Moles of H2O = 2.57 mol x (6/3) ≈ 5.14 mol
5. Finally, calculate the mass of H2O:
Mass of H2O = Moles of H2O x Molar mass of H2O
Since the molar mass of H2O is roughly 18.02 g/mol, we can calculate:
Mass of H2O = 5.14 mol x 18.02 g/mol ≈ 92.57 g
So, to produce 150 grams of Mg(OH)2, approximately 92.57 grams of water are needed.