A mixture of two liquids, hydrazine N2H4 and N2O4 are used in rockets. They produce N2 and water vapors. How many grams of N2 gas will be formed by reacting 100g of N2H4 and 200g of N2O4
2N2H4+N2O4 3N2+4H2O
Answers
Answer:
60.79g
Explaination:
2N2H4 + N204 → 3N2 + 4H20
2N2H4 + N204 → 3N2 + 4H20First step:
2N2H4 + N204 → 3N2 + 4H20First step: Converting the mass of N2H4 and N204 into moles.
2N2H4 + N204 → 3N2 + 4H20First step: Converting the mass of N2H4 and N204 into moles.mass of N2H4 = 100g
2N2H4 + N204 → 3N2 + 4H20First step: Converting the mass of N2H4 and N204 into moles.mass of N2H4 = 100gmolar mass of N2H4= 14x2 + 1.008x4 =32.032 gmol-1
2N2H4 + N204 → 3N2 + 4H20First step: Converting the mass of N2H4 and N204 into moles.mass of N2H4 = 100gmolar mass of N2H4= 14x2 + 1.008x4 =32.032 gmol-1moles of N2H4 = mass in g/ molar mass = 100g/32.032 = 3.12mo
2N2H4 + N204 → 3N2 + 4H20First step: Converting the mass of N2H4 and N204 into moles.mass of N2H4 = 100gmolar mass of N2H4= 14x2 + 1.008x4 =32.032 gmol-1moles of N2H4 = mass in g/ molar mass = 100g/32.032 = 3.12moSimilarly, calculating the moles of N204 which is 2.17
2N2H4 + N204 → 3N2 + 4H20First step: Converting the mass of N2H4 and N204 into moles.mass of N2H4 = 100gmolar mass of N2H4= 14x2 + 1.008x4 =32.032 gmol-1moles of N2H4 = mass in g/ molar mass = 100g/32.032 = 3.12moSimilarly, calculating the moles of N204 which is 2.17Second step:
2N2H4 + N204 → 3N2 + 4H20First step: Converting the mass of N2H4 and N204 into moles.mass of N2H4 = 100gmolar mass of N2H4= 14x2 + 1.008x4 =32.032 gmol-1moles of N2H4 = mass in g/ molar mass = 100g/32.032 = 3.12moSimilarly, calculating the moles of N204 which is 2.17Second step:Comparing the moles of N2H4 and N2
2N2H4 + N204 → 3N2 + 4H20First step: Converting the mass of N2H4 and N204 into moles.mass of N2H4 = 100gmolar mass of N2H4= 14x2 + 1.008x4 =32.032 gmol-1moles of N2H4 = mass in g/ molar mass = 100g/32.032 = 3.12moSimilarly, calculating the moles of N204 which is 2.17Second step:Comparing the moles of N2H4 and N2N204 : N2
2N2H4 + N204 → 3N2 + 4H20First step: Converting the mass of N2H4 and N204 into moles.mass of N2H4 = 100gmolar mass of N2H4= 14x2 + 1.008x4 =32.032 gmol-1moles of N2H4 = mass in g/ molar mass = 100g/32.032 = 3.12moSimilarly, calculating the moles of N204 which is 2.17Second step:Comparing the moles of N2H4 and N2N204 : N22 3
2N2H4 + N204 → 3N2 + 4H20First step: Converting the mass of N2H4 and N204 into moles.mass of N2H4 = 100gmolar mass of N2H4= 14x2 + 1.008x4 =32.032 gmol-1moles of N2H4 = mass in g/ molar mass = 100g/32.032 = 3.12moSimilarly, calculating the moles of N204 which is 2.17Second step:Comparing the moles of N2H4 and N2N204 : N22 32/2 = 1 3/2
2N2H4 + N204 → 3N2 + 4H20First step: Converting the mass of N2H4 and N204 into moles.mass of N2H4 = 100gmolar mass of N2H4= 14x2 + 1.008x4 =32.032 gmol-1moles of N2H4 = mass in g/ molar mass = 100g/32.032 = 3.12moSimilarly, calculating the moles of N204 which is 2.17Second step:Comparing the moles of N2H4 and N2N204 : N22 32/2 = 1 3/2 3.12 3/2 x 3.12 = 4.68
2N2H4 + N204 → 3N2 + 4H20First step: Converting the mass of N2H4 and N204 into moles.mass of N2H4 = 100gmolar mass of N2H4= 14x2 + 1.008x4 =32.032 gmol-1moles of N2H4 = mass in g/ molar mass = 100g/32.032 = 3.12moSimilarly, calculating the moles of N204 which is 2.17Second step:Comparing the moles of N2H4 and N2N204 : N22 32/2 = 1 3/2 3.12 3/2 x 3.12 = 4.68Similarly comparing the moles of N204 and N2 we get 2.17 moles that means N204 is the limiting reactant and it produces 2.17 moles of N2
2N2H4 + N204 → 3N2 + 4H20First step: Converting the mass of N2H4 and N204 into moles.mass of N2H4 = 100gmolar mass of N2H4= 14x2 + 1.008x4 =32.032 gmol-1moles of N2H4 = mass in g/ molar mass = 100g/32.032 = 3.12moSimilarly, calculating the moles of N204 which is 2.17Second step:Comparing the moles of N2H4 and N2N204 : N22 32/2 = 1 3/2 3.12 3/2 x 3.12 = 4.68Similarly comparing the moles of N204 and N2 we get 2.17 moles that means N204 is the limiting reactant and it produces 2.17 moles of N2Now, calculating the mass in grams of N2
moles of N2 = 2.17
moles of N2 = 2.17molar mass = 28
moles of N2 = 2.17molar mass = 28Formula : Mass = moles x molar mass = 2.17 x 28 = 60.76g
moles of N2 = 2.17molar mass = 28Formula : Mass = moles x molar mass = 2.17 x 28 = 60.76g