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.
Please help and explain how to solve this so I may understand it better please and thank you.
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
To solve this problem, you need to identify the limiting reactant, which is the reactant that will be completely consumed and determines the maximum amount of product that can be formed. In this case, the limiting reactant is HCl.
To determine the limiting reactant, compare the mole ratio of the limiting reactant (HCl) to the other reactant (NaOH) using the balanced equation:
HCl + NaOH → NaCl + H2O
The mole ratio between HCl and NaCl is 1:1. This means that for every 1 mole of HCl, 1 mole of NaCl will be formed.
Given that we have 1.2 moles of HCl, we can conclude that 1.2 moles of NaCl will be formed as well.