Question

Uuel Score: 1.50/ Autos Tutored Practice Problem 8.2.4 COUNTS TOWARDS GRADE Use reaction stoichiometry to calculate amounts of reactants and products. Close Problem The substances magnesium nitride and water react to form magnesium hydroxide and ammonia. Unbalanced equation: Mg3N2 (5) + H20 (1) Mg(OH)2 (aq) + NH3 (aq) In one reaction, 66.4 g of NH, is produced. What amount (in mol) of H20 was consumed? What mass (in grams) of Mg(OH)2 is produced? mol H2O consumed g Mg(OH)2 produced Check & Submit Answer Show Approach

Answer 1

I Given Unbalanced equation: Mg ₂ N2 (5) + H₂O(l) mg (OH)₂ (ar) + Mtz (aq) We have at the to balance the equation to know how reactants give how many moles of the many moles products. Mg₃N2 (s) + 6H₂O(e) = 3Mg(OH), (aq) + 2NH₂ (aq) Balanced Ghaeme Now, 66.4 g mtz 5 equation) (equal sign is used for balanced equation). 66.49 - 3.9 mol Mtz is 17 g/mol produced the above balanced equation 06.4 g mtz = Now, from for 2 mol of MHz 6 mol of H₂O is consumed.. so, for, 3.9 mol of m X ) = 11.7 mol of H₂O is Answer I consumed. Again, 6 mol of H₂O, produces 3 mol of Mg(OH)2 Then, 11•7 mol of H2 0 , pooduces, .) = 5.85 mol of Mg(OH)2 5.85 mol x 58 g/moll of Mg(OH)2 Answer II - 339.3 g of mg (Olt), * * Molar mass of Mtz - 14+3 = 17 g/mol molar mass of Mg(OH)₂ = {24 + 2 (16+1)} = (24 +34) * = 58 g/mol

Answers :

11.7 mol H2O consumed.

339.3 g Mg(OH)2  produced.

[** Note : If you are provided with the molar masses of the mentioned compounds, then please use those values to obtain the answers. Molar masses I've used here, are the nearest integer values.]