1.9 mol HCl and 3.8 mol NaOH react according to the equation

HCl + NaOH −→ NaCl + H2O .
Calculate the amount in moles of NaCl
formed.
Answer in units of mol

2 answers

This is limiting reagent problem.
1.9 mol HCl can form 1.9 mol NaCl if it has all of the NaOH it needs (It will need 1.8 mol NaOH).
3.8 mol NaOH can form 3.8 mol NaCl if it has all of the HCl needed(it would need, of course, 3.8 mol HCl. It doesn't have that much).
Therefore, the 1.9 mol HCl will form 1.9 mol NaCl and it will use 1.9 mol NaOH in the process. There will be some NaOH remaining unreacted. 3.8-1.9 = 1.9 moll NaOH unreacted.
15.4
Similar Questions
  1. suppose 23mL of NaOH was required. How much Carbonic Acid was neutralized?1L of .01 NaOH contains ____ mol NaOH? lmL of .01 NaOH
    1. answers icon 0 answers
  2. mol KHP = (0.7719g)(1 mol KHP/204.2g) = .0038molWhere are the moles of NaOH? I know that there are 0.02637 L of NaOH and I know
    1. answers icon 1 answer
    1. answers icon 1 answer
  3. NaOH(s)+ H2SO4(aq)=Na2SO4(aq) + H2O(l)Consider the unbalanced equation above.A 0.900 g sample of impure NaOH was dissolved in
    1. answers icon 2 answers
more similar questions