All questions are related each other therefore I posted all together. I balanced
ID: 906385 • Letter: A
Question
All questions are related each other therefore I posted all together. I balanced the equation but I still need help on question 2 to 5. thank you
1) Hydrogen cyanide is produced industrially from the reaction of gaseous ammonia, oxygen an methane. Balance the reaction for the production of hydrogen cyanide:
2NH3(g) + 3O2(g) + 2CH4(g) 2HCN(g) + 3H2O(g)
2) Using the balanced reaction for production of hydrogen cyanide above, if 3.84 moles of methane are reacted, how many moles of oxygen will be consumed? How many moles of water will be produced?
3) Using the balanced reaction for production of hydrogen cyanide above, if 25.1 g of ammonia are reacted (assume all other reagents in excess), how many grams of hydrogen cyanide will be produced?
4) If I wish to perform a hydrogen cyanide synthesis reaction (using the balanced equation you found for question 1), and react 48.3 g ammonia with 56.8 g oxygen and 32.4 g methane, what masses of hydrogen cyanide and water will be produced?
5) If the reaction in question 4 is performed (using the amounts of reactants listed), but only 27.2 grams of hydrogen cyanide are actually produced, what is the percent yield of the reaction?
Explanation / Answer
(1)The balanced equation is : 2NH3(g) + 3O2(g) + 2CH4(g) 2HCN(g) + 6H2O(g)
Molar mass of NH3 = 14 + (3x1) = 17 g/mol
Molar mass of O2 = 2x16 = 32 g/mol
Molar mass of CH4 = 12 + (4x1) = 16 g/mol
Molar mass of HCN = 1+12+14=27 g/mol
Molar mass of H2O = (2x1) +16 = 18 g/mol
2) 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
(3) 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
(4) 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
(5) Actual mass of HCN = 27.2 g
Theoretical mass of HCN = 31.95 g
So percent yield=(actual mass / thoretical mass ) x100
= ( 27.2 / 31.95 ) x100
= 85.1 %