Consider the reaction represented by the following equation: Na₂CO_3(aq) + 2HCl_(aq) \(\longrightarrow\) 2NaCl_(aq) + H₂O_(l) + CO_2(g) . What volume of 0.02 mol dm^-3 Na₂CO_3(aq) would be required to completely neutralize 40 cm³ of 0.10 mol dm^-3 HCl_(aq)?

200 cm³

100 cm³

40 cm³

20 cm³

Correct Option

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10 years ago

CaVa/CbVb equals Na/Nb

40*0.10/(0.02*Vb) =2/1

therefore Vb=100Cm3

Answer B

10 years ago

how come?

10 years ago

Ca*Va/Cb*Vb=Na/Nb .1*40/.02*V=2/1 = 100cm3

10 years ago

d correct answer is 100cm3 from caba/cbvb=na/nb

10 years ago

Thanks for your contributions. Correction has been made.

5 years ago

CA=0.02 mol dm-3,
VA= ?, nA=2
CB= 0.10 mol dm-3,
VB= (CB VB NA) / (CA NB)
= (0.10×40×2) / (0.2×1)
= 40cm3
.: The answer is C

ceoofwahala

20th June, 2026

Chemistry

2 comments

ASSAAS

20th June, 2026

English Language

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infinitehoaxx

21st May, 2026

Computer

4 comments

Consider the reaction represented by the following equation: Na2CO3(aq) + 2HCl(aq) \(\longrightarrow\) 2NaCl(aq) + H2O(l) + CO2(g) . What volume of 0.02 mol dm-3 Na2CO3(aq) would be required to completely neutralize 40 cm3 of 0.10 mol dm-3 HCl(aq)?