chemistry

1.2 mol HCl and 3.9 mol NaOH react to this equation,
HCl + NaOH=NaCl + H20
If the limited reactant id HCl, calculate the amount of NaCl formed in moles.

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1. I = initial; i.e., start
C = change
E = equilibrium; end

.......HCl + NaOH ==> NaCl + H2O
I......1.2...3.9.......0......0
C.....-1.2..-1.2......+1.2...+1.2
E......0.....2.7.......1.2....1.2

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