Question

how many grams of sodium hydroxide is required to neutralise 73 grams of hydrochloric acid and to form sodium chloride and water​

Answer 1

Answer:

The balanced reaction is as follows:

2Na + 2H2O → 2NaOH + H2

Moles of metallic sodium =

0.46/23 =0.02

Moles of HCl= Moles of NaOH

Molarity × Volume = Number of moles

73/36.5×1 ×V =0.02

⟹V=10L

V = 10 × 1000

= 10000g

Hence, volume of the solution required would be 10000g

Answer 2

Answer:

80g

Explanation:

1 mole of sodium hydroxide requires 1 mole of hydrochloric acid

So 2 moles of sodium hydroxide is required to neutralise 2 moles of hydrochloric acid[73/36.5=2]

1 mole weighs 40g so 2 moles weigh 80g

Mark me as brainliest if you wish

Hope it helps

If you want the equation

NaOH+HCl=NaCl+H2O