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
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
To determine the amount of hydrochloric acid (HCl) needed to react with 40g of sodium hydroxide (NaOH), we need to write a balanced chemical equation for the reaction between HCl and NaOH.
The balanced equation for this reaction is:
HCl + NaOH → NaCl + H2O
To balance this equation, we need one molecule of HCl and one molecule of NaOH to react, which will produce one molecule of NaCl and one molecule of H2O.
Now, let's calculate the molar masses of the compounds involved:
- HCl: 1g/mol (hydrogen: 1g/mol + chlorine: 35.5g/mol)
- NaOH: 40g/mol (sodium: 23g/mol + oxygen: 16g/mol + hydrogen: 1g/mol)
- NaCl: 58.5g/mol (sodium: 23g/mol + chlorine: 35.5g/mol)
- H2O: 18g/mol (hydrogen: 1g/mol + oxygen: 16g/mol)
To find the required amount of HCl to react with 40g of NaOH, we need to determine the molar ratio between HCl and NaOH using their molar masses.
Molar ratio:
(1 mol of HCl / 1 g) = (40 g of NaOH / X mol)
X = (40 g of NaOH) / (molar mass of NaOH)
X = 40g / 40g/mol = 1 mol
So, 1 mol of HCl is required to react with 40g of NaOH.
The molar mass of HCl is 36.5g/mol (hydrogen: 1g/mol + chlorine: 35.5g/mol). Therefore, you should use 36.5g of HCl for the reaction with 40g of NaOH.
Now let's calculate the mass of NaCl and H2O produced:
- 1 mol of NaCl is produced for every 1 mol of HCl used.
- 1 mol of HCl weighs 36.5g, and 1 mol of NaCl weighs 58.5g.
- Therefore, the mass of NaCl produced would be 58.5g.
- Also, 1 mol of NaOH reacts with 1 mol of HCl and produces 1 mol of H2O.
- 1 mol of H2O weighs 18g.
- So, the mass of H2O produced would be 18g.
In conclusion:
- 36.5g of HCl is needed to react with 40g of NaOH.
- 58.5g of NaCl will be produced.
- 18g of H2O will be produced.