Question

A solution of NaOH is 4 g L-1. What volume of HCI
gas at STP will neutralize, 50 ml of the alkali
solution?
(1) 224 ml
(2) 112 ml
3) 11200 ml
(4) 22.4 ml​

Answer 1

Answer:

Therefore, the volume of HCl gas at STP will be, 0.112 L or 112 ml.

Explanation:

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Answer 2

Answer:

The balanced chemical reaction is

NaOH + HCl → NaCl + H₂O

In 1 Litre, the NaoH = 4gm

or in 1000 ml, the NaOH = 4 gm

Therefore in 50 ml, NaoH = 0.2 gm

Moles of NaOH = Weight (gm)/Molecular Weight

= 0.2/40 = 0.005 mole

From the reation

1 mole of NaOH will neutralise 1 mole of HCl

Therefore 0.005 mole of NaCl will neutralise 0.005 mole of HCl

Volume of 1 mole of HCl is 22.4 litres at STP

Therefore volume of 0.005 moles of HCl is 0.005 x 22.4

= 0.112 Litre = 112 mL