Home
Class 12
CHEMISTRY
When 22.4 L of H(2)(g) is mixed with 11....

When `22.4 L` of `H_(2)(g)` is mixed with 11.2 of `Cl_(2)(g)`, each at STP, the moles of HCl(g) formed is equal to

A

1.5 mol of HCl (g)

B

1 mol of HCl (g)

C

2 mol of HCl (g)

D

0.5 mol of HCl (g)

Text Solution

Verified by Experts

The correct Answer is:
B
`{:(H_2 ,+, Cl_2 ,to, 2HCl),(1 lit ,,1 lit , , 2lit):}`
initial 22.4 lit ` to` 11.2 lit ` to` limiting reagent
`:.` 11.2 lit ` to ` 11.2 lit ` to` 22.4 lit
` :.` number of moles `=(22.4)/(22.4)`=1 mol
Promotional Banner

Topper's Solved these Questions

  • POLYMERS

    NARAYNA|Exercise EXERCISE -4|8 Videos

  • POLYMERS

    NARAYNA|Exercise EXERCISE -2 (H.W.) ( COMMERCIALLY IMPORTANT POLYMERS )|2 Videos

  • PERIODIC TABLE

    NARAYNA|Exercise All Questions|573 Videos

  • PRACTICAL ORGANIC CHEMISTRY

    NARAYNA|Exercise SUBJECTIVE_TYPE|1 Videos

Similar Questions

Explore conceptually related problems

What is the increase in entropy when 11.2 L of O_(2) are mixed with 11.2 L of H_(2) at STP?

In the reaction vessel 100g H_(2) and 100g Cl_(2) are mixed and suitbale conditions are provided for the following reactions: H_(2)(g)+Cl_(2)(g) overset(?)to 2HCl(g) The actual amount of HCl formed in this reaction is:

Consider the reaction at 300 K H_(2)(g)+Cl_(2)(g)to2HCl(g), DeltaH^(@)=-185kJ If 2 mole of H_(2) compeletely react with 2 mole of Cl_(2) to form HCl . What is DeltaU^(@) for this reaction ?