2NH3(g) +3O2(g) +2CH4(g) 2HCN(g) + 6H2O(g)2NH3(g) +3O2(g) +2CH4(g) 2HCN(g) + 6H2
ID: 493554 • Letter: 2
Question
2NH3(g) +3O2(g) +2CH4(g) 2HCN(g) + 6H2O(g)2NH3(g) +3O2(g) +2CH4(g) 2HCN(g) + 6H2O(g)
If 3.84 moles of methane are consumed in this reaction, how many moles of HCN will be produced?
If 25.1 g of ammonia are reacted, how many grams of HCN will be produced?
If I react 48.3 g of ammonia with 32.4 g of methane (assume oxygen is present in excess), what is the maximum mass of HCN that can be produced?
If I react 17.9 g of ammonia with excess oxygen and methane, but only collect 23.3 g of HCN, what is the percent yield?
Explanation / Answer
1) 2NH3(g) + 3O2(g) + 2CH4(g) 2HCN(g) + 6H2O(g)
From the balanced equation,
2 moles of methane reacts with 3 moles of O2
3.84 moles of methane reacts with M moles of O2
M = (3x3.84) / 2
= 5.76 mol
2 moles of methane produces 6 moles of water
3.84 moles of metahne produces N moles of water
N = (6x3.84) /2
= 11.52 mol
(2) 2NH3(g) + 3O2(g) + 2CH4(g) 2HCN(g) + 6H2O(g)
2x17 g of ammonia produces 2x27 g of HCN
25.1 g of ammonia produces Y g of HCN
Y = (25.1x2x27) / (2x17)
= 39.9 g
(3) 2NH3(g) + 3O2(g) + 2CH4(g) 2HCN(g) + 6H2O(g)
2x17 g of NH3 reacts with 3x32 g of O2 & 2x16 g of CH4
34 g of NH3 reacts with 96 g of O2 & 32 g of CH4
M g of NH3 reacts with 56.8 g of O2 & N g of CH4
M = (56.8x34)/96 = 20.12 g of NH3
N = ( 56.8x32) / 96 = 18.9 g of CH4
So all the mass of O2 completly reacted O2 is the limiting reactant.
From the balanced equation ,2NH3(g) + 3O2(g) + 2CH4(g) 2HCN(g) + 6H2O(g)
3x32=96 g of O2 produces 2x27=54 g of HCN and 6x18=108 g of water
56.8 g of O2 produces Y g of HCN and Z g of water
Y = (56.8x54)/96 = 31.95 g of HCN
Z = ( 56.8x108) / 96
= 63.9 g
(4) Actual mass of HCN = 23.3 g
Theoretical mass of HCN = 31.95 g
So percent yield=(actual mass / thoretical mass ) x100
= ( 23.3 / 31.95 ) x100
= 72.92%
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.