Question

When 22.4L of H2g is mixed with 11.2 of cl2 g each at STP the moles of stp the moles of hcl g formed is equal to

Answer 1

This is the question of limiting reactant
22.4 liter of H2 gives 22.4 liter of hcl


And 22.4 liter of cl2 gives 22.4 liter of hcl
1 liter gives 22.4 ÷22.4 liter of hcl
11.2 liter of hcl gives 22.4 ÷ 22.4 ×11.2 liter of hcl
= 11.2 liter of hcl


So cl2 is limiting reactant
11.2 liter of cl2 gives 11.2 liter of hcl
Moles = volume ÷ 22.4
Moles = 11.2 ÷22.4
Moles = 0.5 moles

And its mass is mass = mole × molecular mass
Mass = 0.5 × 36.5
Mass in gram = 18.25

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