56 g of Nitrogen reacts with 22 g of hydrogen to form ammonia.Find
I) The limiting reagent of the reaction.
2) The amount of reactant in excess
3) The amount of ammonia formed.
Answers
Answer:
56g of N2+10g of H2
56g of N2+10g of H2N2(g)+3H2(g)→2NH3(g)
56g of N2+10g of H2N2(g)+3H2(g)→2NH3(g) In the given reaction, 1 mol of N2 reequires 3 moles of H2 for the formation fo ammonia.
56g of N2+10g of H2N2(g)+3H2(g)→2NH3(g) In the given reaction, 1 mol of N2 reequires 3 moles of H2 for the formation fo ammonia.Thus if the number of moles of N2 and H2 should be in ratio 1:3.
56g of N2+10g of H2N2(g)+3H2(g)→2NH3(g) In the given reaction, 1 mol of N2 reequires 3 moles of H2 for the formation fo ammonia.Thus if the number of moles of N2 and H2 should be in ratio 1:3.We know,
56g of N2+10g of H2N2(g)+3H2(g)→2NH3(g) In the given reaction, 1 mol of N2 reequires 3 moles of H2 for the formation fo ammonia.Thus if the number of moles of N2 and H2 should be in ratio 1:3.We know,n=molecular weightweight
56g of N2+10g of H2N2(g)+3H2(g)→2NH3(g) In the given reaction, 1 mol of N2 reequires 3 moles of H2 for the formation fo ammonia.Thus if the number of moles of N2 and H2 should be in ratio 1:3.We know,n=molecular weightweightThus in option C 56g of N2 means 2 moles of N2 and 10g of H2 means 5 moles of H2.
56g of N2+10g of H2N2(g)+3H2(g)→2NH3(g) In the given reaction, 1 mol of N2 reequires 3 moles of H2 for the formation fo ammonia.Thus if the number of moles of N2 and H2 should be in ratio 1:3.We know,n=molecular weightweightThus in option C 56g of N2 means 2 moles of N2 and 10g of H2 means 5 moles of H2.Thus here H2 acts as limiting reagent
Explanation:
The limiting reagent of the reaction :
➲ The limiting reagent of the reaction is Nitrogen
2) The amount of reactant in excess :
➲ The amount of reactant in excess is Hydrogen
3) The amount of ammonia formed :
➲ The amount of ammonia formed is 68 grams