A dirty and impure sample that weighs 43.4 g contains a certain percentage of magnesium nitride. Magnesium nitride reacts with water.The sample is exposed to an excess of water to give 32.9 g Mg(OH)2. What percentage of Mg3N2is in the original sample?

Question

Wyatt Atkins · Accepted Answer

The sample contains 43.7 % #"Mg"_3"N"_2#.

Step1. Calculate the moles of #"Mg"# #"Moles of Mg" = 32.9 color(red)(cancel(color(black)("g Mg(OH)"_2))) × (1 color(red)(cancel(color(black)("molMg(OH)"_2))))/(58.32 color(red)(cancel(color(black)("g Mg(OH)"_2)))) × "1 mol Mg"/(1 color(red)(cancel(color(black)("mol Mg(OH)"_2)))) = "0.5641 mol Mg"# Step 2. Calculate the moles of #"Mg"_3"N"_2# #"Moles of Mg"_3"N"_2 = 0.5641 color(red)(cancel(color(black)("mol Mg"))) × ("1 mol Mg"_3"N"_2)/(3 color(red)(cancel(color(black)("mol Mg")))) = "0.1880 mol Mg"_3"N"_2# Step 3. Calculate the mass of #"Mg"_3"N"_2# #"Mass of Mg"_3"N"_2 = 0.1880 color(red)(cancel(color(black)("mol Mg"_3"N"_2))) × (100.93 "g Mg"_3"N"_2)/(1 color(red)(cancel(color(black)("mol Mg"_3"N"_2)))) = "18.98 g Mg"_3"N"_2# Step 4. Calculate the percent of #"Mg"_3"N"_2# #"% Mg"_3"N"_2 = (18.98 color(red)(cancel(color(black)("g"))))/(43.4 color(red)(cancel(color(black)("g")))) × 100 % = "43.7 %"#

Kai Bowman · Answer

undefined To find the percentage of magnesium nitride (Mg3N2) in the original sample, you can follow these steps:

1. Determine the amount of magnesium hydroxide (Mg(OH)2) produced in the reaction.
2. Use stoichiometry to find the amount of magnesium nitride (Mg3N2) that reacted to produce the magnesium hydroxide.
3. Calculate the percentage of magnesium nitride in the original sample.

Step 1: Calculate the amount of magnesium hydroxide produced:
Mass of Mg(OH)2 = 32.9 g

Step 2: Use stoichiometry to find the amount of magnesium nitride reacted:
The balanced chemical equation for the reaction between magnesium nitride and water is:
Mg3N2 + 6H2O → 3Mg(OH)2 + 2NH3

From the equation, 1 mole of Mg3N2 produces 3 moles of Mg(OH)2.
So, the molar ratio of Mg3N2 to Mg(OH)2 is 1:3.

Calculate the moles of Mg(OH)2:
Moles of Mg(OH)2 = Mass / Molar mass
Moles of Mg(OH)2 = 32.9 g / (24.31 g/mol + 2 * 16.00 g/mol)
Moles of Mg(OH)2 ≈ 0.675 moles

According to the stoichiometry:
1 mole of Mg(OH)2 corresponds to 1/3 mole of Mg3N2.

Moles of Mg3N2 = Moles of Mg(OH)2 / 3
Moles of Mg3N2 ≈ 0.675 moles / 3
Moles of Mg3N2 ≈ 0.225 moles

Step 3: Calculate the percentage of Mg3N2 in the original sample:
Mass of Mg3N2 = Moles of Mg3N2 × Molar mass of Mg3N2
Mass of Mg3N2 = 0.225 moles × (3 * 24.31 g/mol + 2 * 14.01 g/mol)
Mass of Mg3N2 ≈ 0.225 moles × 100.77 g/mol
Mass of Mg3N2 ≈ 22.68 g

Percentage of Mg3N2 in the original sample = (Mass of Mg3N2 / Mass of original sample) × 100%
Percentage of Mg3N2 in the original sample = (22.68 g / 43.4 g) × 100%
Percentage of Mg3N2 in the original sample ≈ 52.25%

Therefore, approximately 52.25% of Mg3N2 is present in the original sample.