HCl (aq) +NaOH (aq) →NaCl (aq) +H 2 O (l) H, C, l, start subscript, left parenthesis, a, q, right parenthesis, end subscript, plus, N, a, O, H, start subscript, left parenthesis, a, q, right parenthesis, end subscript, right arrow, N, a, C, l, start subscript, left parenthesis, a, q, right parenthesis, end subscript, plus, H, start subscript, 2, end subscript, O, start subscript, left parenthesis, l, right parenthesis, end subscript3.4 moles of HCl react with 3.2 moles of NaOH. How many moles of HCl remain when the reaction stops?
Question
HCl (aq) +NaOH (aq) →NaCl (aq) +H 2 O (l) H, C, l, start subscript, left parenthesis, a, q, right parenthesis, end subscript, plus, N, a, O, H, start subscript, left parenthesis, a, q, right parenthesis, end subscript, right arrow, N, a, C, l, start subscript, left parenthesis, a, q, right parenthesis, end subscript, plus, H, start subscript, 2, end subscript, O, start subscript, left parenthesis, l, right parenthesis, end subscript3.4 moles of HCl react with 3.2 moles of NaOH. How many moles of HCl remain when the reaction stops?
Solution
The reaction between HCl and NaOH is a neutralization reaction, where one mole of HCl reacts with one mole of NaOH to produce one mole of NaCl and one mole of H2O. This is a 1:1:1:1 stoichiometric ratio.
Given that 3.4 moles of HCl react with 3.2 moles of NaOH, the limiting reactant is NaOH because it is present in a lesser amount. This means all of the NaOH will be used up in the reaction and there will be some HCl left over.
To find out how many moles of HCl remain, we subtract the number of moles of NaOH used (which is also the number of moles of HCl used) from the total number of moles of HCl.
So, 3.4 moles HCl - 3.2 moles NaOH = 0.2 moles HCl remain when the reaction stops.
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