Yesterday we combined hydrochloric acid hcl with sodium hydroxide naoh in a violent reaction that resulted in water h2o and common table salt nacl.
how many grams of hydrochloric acid should we use so we have exactly enough to react with 40g of sodium hydroxide? you'll need to write a chemical reaction, balance it, and then perform your calculation.
also how many grams of salt nacl will be produced and how many grams of water h2o be produced?

Reaction:
HCl + NaOH ---> NaCl + H2O

1 mole of HCl = 36,5 g
1 mole of NaOH = 40g

so, according to the reaction:
1 mol HCl = 1 mol NaOH
so, we need > 36,5 g HCl (hydrochloric acid)

36,5 g HCl (hydrochloric acid)
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next question.

1 mole of NaCl = 58,5 g
1 mole of H2O = 18g

so, according to the reaction:
1 mole of HCl (36,5 g) - 1 mole of NaCl (58,5 g)
(the same for NaOH)
i
1 mole of HCl (36,5 g) 1 mole of H2O (18 g)
(the same for NaOH)

so, this reaction is stechiometric

58,5 g NaCl i 18g H2O

HCl + NaOH ---> NaCl + H2O                  

                                                        1 mol HCl = 1 mol NaOH 

mole HCl = 36.5 g // mole HaOH = 40 g

36.5 HCl is the answer.


mole NaCl = 58,5 g // mole H2O = 18g 


HCl (36,5 g) ⇒ NaCl (58,5 g)
HCl (36,5 g) ⇒ H2O (18 g)

So, 58,5 g NaCl and 18 g H2O are the answer.

It will appear as o2(g) as a reactant