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AP Chemistry molar mass mixture problem?
Consider the following two reactions, each of which will go to completion on strong heating:
Mg(OH)2(s) --> MgO(s) + H2O(g)
Mg3N2(s) + 3H2O(l) --> 3MgO(s) + 2NH3(g)
If a 2.50 g mixture of Mg(OH)2(s) and Mg3N2(s) is treated with water and strongly heated giving 2.00 g
MgO as the exclusive solid product, what was the mass of Mg(OH)2(s) in the starting mixture?
1 Answer
- Roger the MoleLv 77 years agoFavorite Answer
Let z be the mass (in grams) of Mg(OH)2 to be found.
Then 2.50-z is the mass of Mg3N2 in the mixture.
z / (58.3197 g Mg(OH)2/mol) x (1 mol MgO / 1 mol Mg(OH)2) x (40.30440 g MgO/mol) =
(0.691094 z) g MgO from Mg(OH)2
(2.50-z) / (100.9284 g Mg3N2/mol) x (3 mol MgO / 1 mol Mg3N2) x (40.30440 g MgO/mol) =
(2.99502 - 1.19801 z) g MgO from Mg3N2
Add the two sources of MgO and set the sum equal to the given amount of MgO produced:
(0.691094 z) + (2.99502 - 1.19801 z) = 2.00
Solve for z algebraically:
z = 1.96 g Mg(OH)2
