50.0kg of N2[g] and 10.0 kg of H2 [g] are mixed to produce NH3[g] formed.Identify the limiting reagent in production of NH3 in this situaton
Asked by virubloda6 | 21st May, 2019, 08:39: AM
Mass of N2 = 50 kg
= 50 × 103 gm
No. of moles of N2
Mass of H2 = 10 kg
= 10 × 103 gm
No. of moles of H2
Formation of ammonia;
From reaction, 1 mole of N2 requires 3 moles H2.
So 1.78× 103 mole of N2 will require (1.78×103 ×3) = 5.34 ×103 moles of H2.
Here 5×103 moles of H2 are given, therefore H2 is the limiting reagent.
Now 5×103 moles of H2 will give,
Therfore the amount of NH3 produced is 3.33×103 mole.
Answered by Varsha | 21st May, 2019, 11:52: AM
- 0.315g of organic compound gave on combustion 0.4131g of water and 0.6732g of carbon dioxide calculate the percentage of carbon and hydrogen
- IS ALL PHYSICAL QUANTITIES WHICH ARE MEASUREBLE ARE MATTER OR NON MATTER?
- 0.38g of NaOH is dissolved in water to prepare 50 ml solution. calculate the molarity of the solution
- molecular mass
- 324g of 20% by mass of Ba(OH)2 solution
- what is molecular mass?
- How many gram of nh3 will be produce when 50g N2 react with h2
- what is mole
- In a reaction of magnesium with HCl if 2.24 litre of H2 is obtained at STP the mass of the other product will be what
- 5g of marble was added to 100 c.c of 0.2 M HCl. What volume of CO2 measured at STP will be evolved in the above reaction
Kindly Sign up for a personalised experience
- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions
Verify mobile number
Enter the OTP sent to your number