Question #234674
Excess dilute HCl was added to sodium sulphite 960cm3 of sulphur IV oxide has was produced calculate the mass of the sulphate used
1
Expert's answer
2021-09-10T00:41:48-0400

Putting molar gas volume =24000cm324000cm^3 and molar mass of sulphite =126g, mass of the sulphite used can be calculated as follows.

number of moles of sulphur (iv) oxide =960cm3×1mole24000cm3=0.04moles=\frac{960cm^3 \times 1mole}{24000cm^3}=0.04moles

Equation:

Na2SO3(s)+2HClaq2NaClaq+SO2(g)+H2O(l)Na_2SO_3 (s)+2HCl_{aq} \to 2NaCl_{aq} +SO_2(g)+H_2O_(l)

mole ration Na2SO3:SO2Na_2SO_3:SO_2 is 1:1

number of moles of Na2SO3=0.04molesNa_2SO_3=0.04moles

mass of Na2SO3=126gmol1×0.04=5.04gNa_2SO_3=126gmol^{-1} \times 0.04=5.04g

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