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How many ml of 0.125 M Na2SO4 are needed to react with excess Pb(NO3)2 to obtain...

How many ml of 0.125 M Na2SO4 are needed to react with excess Pb(NO3)2 to obtain 0.750 grams of solid PbSO4?
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Homework Answers

Answer #1

the reaction is

Na2SO4 + Pb (NO3)2 PbSO4 + 2NaNO3

from the reaction,

1 moles of sodium sulphate produces 1 mole of lead sulphate.

ie.,142.04 grams/mole of sodium sulphate produces 303.26 grams/mole of lead sulphate

then 0.750 grams of PbSO4 is formed from (142.04 * 0.750) / 303.26 = 0.351 grams of sodium sulphate.

therefore number of moles of sodium sulphate = 0.351gr / 142.04 = 0.00247moles

we know

molarity M = number of moles / volume in lit

ie.., 0.125 M = 0.00247 moles / volume

therefore volume = 0.00247 / 0.125 = 0.019785 lit = 19.785ml of sodium sulphate is needed

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