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

When 22.4L of H2g is mixed with 11.2 of cl2 g each at STP the moles of

1.	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» 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 HOPE It helps you ... Markas brainlist plz PlzZ