When 10.3 mol na react with 11.5 mol hcl, what is the limiting reactant and how many moles of h2 can be formed? unbalanced equation: na + hcl nacl + h2 na is the limiting reactant, and 5.15 mol h2 can be formed hcl is the limiting reactant, and 5.75 mol h2 can be formed na is the limiting reactant, and 3.43 mol h2 can be formed hcl is the limiting reactant, and 3.83 mol h2 can be formed?

Question

Teagan Archer

Chemistry

18 October, 05:10

When 10.3 mol na react with 11.5 mol hcl, what is the limiting reactant and how many moles of h2 can be formed? unbalanced equation: na + hcl nacl + h2 na is the limiting reactant, and 5.15 mol h2 can be formed hcl is the limiting reactant, and 5.75 mol h2 can be formed na is the limiting reactant, and 3.43 mol h2 can be formed hcl is the limiting reactant, and 3.83 mol h2 can be formed?

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Javier Downs · Answer 1

The balanced equation tells us that Na and HCl react in a 2:2 mole ratio, which is the same as a 1:1 ratio. So if we have 10.3 moles of Na and 11.5 moles of HCl, the Na will be used up first since there's less of it. So Na is the limiting reactant and will determine how much H2 is produced.

The balanced equation also tells us that 2 moles of Na produce 1 mole of H2, so

10.3 moles Na x (1 mole H2 / 2 moles Na) = 5.15 moles H2 produced.