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.
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Answers
Explanation:
Answer :-
1) Limiting reagent is Nitrogen .
2) Hydrogen is present in excess .
3) 68 grams of ammonia is formed .
Explanation :-
The balanced chemical equation for the reaction is :-
N₂ + 3H₂ → 2NH₃
Moles of N₂ :-
= Given Mass/Molar mass
= 56/28
= 2 moles
Moles of H₂ :-
= Given Mass/Molar mass
= 22/2
= 11 moles
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From the balanced equation, we can understand that 1 mole of N₂ will combine with 3 moles of H₂ . Then, 2 moles of N₂ will combine with 6 moles of H₂ .
So, nitrogen (N₂) is the limiting reagent and it will control the amount of product .
Hydrogen (H₂) is present in excess .
________________________________
Now, again from the balanced equation :-
∵ 28 g of N₂ → 34 g of NH₃
∴ 56 g of N₂ will give :-
= 56 × 34/28
= 2 × 34
= 68 g of NH₃.
Answer:
To find : The distance between the school and his house
Solution :
Speed of a student = 2 ½ km/h = 5/2km/h
★ A student reaches his school 6 minutes late.
Consider the time be x
60min = 1 hour
Time = (x + 6) min = (x + 6/60) = (x + 1/10)h
As we know that
★ Speed = distance/time
Consider the distance be y
\implies \sf \dfrac{5}{2} = \dfrac{y}{ x + \dfrac{1}{10}}⟹25=x+101y
\implies \sf \dfrac{5}{2} = \dfrac{y}{ \dfrac{10x + 1}{10} }⟹25=1010x+1y
\implies \sf \dfrac{5}{2} = y \times \dfrac{10}{10x +1}⟹25=y×10x+110
\implies \sf \dfrac{5}{2} = \dfrac{10y}{10x +1}⟹25=10x+110y
\implies \sf 5(10x +1) = 20y⟹5(10x+1)=20y
\implies \sf 50x +5=20y⟹50x+5=20y
\implies \sf 50x - 20y = - 5 \: \: \: \bf(Equation \: 1)⟹50x−20y=−5(Equation1)
★ Next day starting at the same time, he increases his speed by 1 km/hour and reaches 6 minutes early.
\implies \sf \dfrac{5}{2} + 1 = \dfrac{y}{ x - \dfrac{1}{10}}⟹25+1=x−101y
\implies \sf \dfrac{5 + 2}{2} = \dfrac{y}{\dfrac{10x - 1}{10}}⟹25+2=1010x−1y
\implies \sf \dfrac{7}{2} = \dfrac{10y}{10x - 1}⟹27=10x−110y
\implies \sf 7(10x - 1) = 20y⟹7(10x−1)=20y
\implies \sf 70x - 20y = 7 \: \: \: \bf(Equation \: 2)⟹70x−20y=7(Equation2)
Subtract both the equations
→ 50x - 20y - (70x - 20y) = -5 - 7
→ 50x - 20y - 70x + 20y = - 12
→ - 20x = - 12
→ x = 12/20 = 3/5
Put the value of x in eqⁿ (1)
→ 50x - 20y = - 5
→ 50 × 3/5 - 20y = - 5
→ 30 - 20y = - 5
→ 30 + 5 = 20y
→ 35 = 20y
→ y = 35/20 = 7/4
→ y = 1 3/4 km
•°• The distance between school and house is 1 3/4 km.