Question

1. Using the following equation:  2NH₃  + 3Cl₂ -->  N₂  + 6HCl

   if you had 5.64 g of Cl₂ gas, how many moles of hydrochloric acid would be produced?

   if you had 0.653 mol of HCl gas produced, how many g of ammonia gas was used?

Answer 1

02) 8N3 (1 + 30/219) - N2 (g) + 6H10 g) mass of cl= 5.64g molar mass of Cl₂ = 71 g moel -) no. of moles of Cl2 = 5,64 = 0.0794 Mole According to above Balanceel reaction, it is the one clear that 3 mole of ce, give = 6 mole of tice I mole of cly give = mole of the + 0.0194 malo of Clz give = € X0,0194 = 0.1588 - mole 1 0.0794 mole of cl give = 0.1588 mole of hice Thus 5.64g op cez produced 0.1588 male of Hydrochloric ...aud. Aus. I we have given = 0.653 mole of lice. Now, from Balanced chemical reachion in Palt (q; it is clear that 6 mole rice produced by = 2 mole of NH3

I = I mole of Hice produced by = 2 mole of why 5) 0.653 malo op Hel produced by= 2 x0.653 mole of KEER N Her =>0.653 Male of Hcl produced by = 0.2177 mole ofrutta thus =) we have mole of NH₃ = 0.2177 mol a molar mass of NH3 = 17 gmał = mass of NH₃ = 0.2177X17 = 3.70099 Thus we can say 7 10.653 mole of Hid gas produced by 3.70099 of NH3 Ang.