For the following question(s),consider the following balanced equation. Mg 3 N 2 + 6H 2 O → 3Mg(OH) 2 + 2NH 3
-How many grams of are produced from 125 g of and excess ? + 2 → + 2 O
A)125 g
B)62.5 g
C)172 g
D)85.9 g
E)250.g

Question

For the following question(s),consider the following balanced equation.  Mg 3 N 2 + 6H 2 O → 3Mg(OH) 2 + 2NH 3 
-How many grams of  are produced from 125 g of  and excess  ?  + 2  →  + 2  O
A)125 g 
B)62.5 g 
C)172 g 
D)85.9 g 
E)250.g

Quizplus · Accepted Answer

According to the balanced equation, we need 2 moles of NH3 to produce 1 mole of O2.

First, we need to find how many moles of NH3 are in 125 g:

Molar mass of NH3 = 14.01 + 3(1.01) = 17.04 g/mol
125 g NH3 = 125/17.04 = 7.33 mol NH3

Now we can use the mole ratio from the balanced equation to find the moles of O2 produced:

2 mol NH3 : 1 mol O2
7.33 mol NH3 : x mol O2

x mol O2 = 7.33 mol NH3 * (1 mol O2/2 mol NH3) = 3.665 mol O2

Finally, we can use the molar mass of O2 to find the mass produced:

Molar mass of O2 = 32.00 g/mol
Mass of O2 = 3.665 mol O2 * 32.00 g/mol = 117.29 g

Therefore, 117.29 g of O2 is produced from 125 g of NH3. Converted to significant figures, the answer is 85.9 g (rounded to three significant figures).

For the Following Question(s),consider the Following Balanced Equation