Question

the volume in ml of 0.1n hcl to react completely with 1gram mixture of na2co3 and nahco3 containing equimolar amounts of the two components is

Answer 1

Answer:

Explanation:

equimolar mixtures means equal moles

suppose grams of na2co3 = x

grams of nahco3 should be 1-x

valency factor of na2co3 = 2

valency factor of nahco3 = 1

moles of  na2co3= moles of nahco3

x/142 = 1-x/ 84

84x = 142 - 142x

226x = 142

x=0.628

equivalent of na2co3 = 88

equivalent of  nahco3 = 44

equivalent of  na2co3 + equivalent of nahco3 = 0.1n

132 = 0.1 n

1320 ml = volume