At STP, if 4.20L of O2 reacts with N2H4, how many liters of water vapor will be produced?
Answers
Answer:
The volume of steam produced when 4.20 L of O2 reacts with N2H4 at stp is 2.8 Liters.
Explanation:
Step 1: Write a balanced equation for the reaction.
N2H4 + 3O2 —> 2 NO2 + 2H2O
Step 2 : Identify the correct formula to use.
Molar gas volume at stp is 22.4 Liters.
Step 3 : Calculate the moles of Oxygen in 4.20 Liters.
1 mole — 22.4 Liters.
The moles in 4.20 Liters is :
4.20/22.4 = 0.1875 moles
Step 4 : Work out the mole ratio between Oxygen and the water vapor produced.
The mole ratio from the equation is = 3 : 2
Step 5 : work out the moles of Steam from the mole ratio.
2/3 × 0.1875 moles = 0.125 moles.
Step 6: workout the volume of steam produced given the moles.
1 mole = 22.4 liters
0.125 moles =?
0.125 × 22.4 = 2.8 Liters.
Answer:
8.40L H2O(g)
Explanation:
4.2L O2 x 1mol o2/22.4L o2 x 2mil h2o/1mol o2 x22.4l o2/1mol o2 =8.40L H2O