Dear Sally,
as I have replied you before, I am not sure whether you have checked
HCL(aq) + NAOH(aq)---- NACL(s)+ H20(l)
Number of moles in the NAOH(aq)=
40/(23+16+1)
=1
as 1 mole of HCL will react with 1 mole of the Naoh solution
number of mole of HCL required= 1
Still remember that there is a equation
mol= g/molar mass
Let the number of grams of HCL needed be g grams
the number of grams of HCL required=
1= g/(1+35.5)
g=36.5
Therefore, 36.5 grams of HCL is required
For the grams of Nacl produced
1 mole of Hcl will produce 1 mole of nacl
Let the number of grams of salt produced be g grams
1x(23+35.5)
=55.5
Therefore 55.5 grams of nacl is produced
For the grams of water produced
1 mole of Hcl will produce 1 mole of water
Let the number of gram of the water produced be g grams
so 1= g/18
g=18
Therefore, 18 grams of water is produced
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?
THANK YOU <3
1 answer
1 answer