Question

Na2 S o4 + pb(no3)2 gives us pbso4 + 2 Nano 3 when excess lead nitrate solution was added to a solution of sodium sulphate 15.15 gram of lead sulphate war precipitated what mass of sodium sulphate was present in the original solution

Answer 1

Given, Na2SO4+Pb(NO3)2→PbSO4+2NaNO3Molar mass of Na2SO4=(23×2+32+16×4) =142gMolar mass of PbSO4=(207+32+16×4) =303g303g of lead sulphate is precipitated from 142g of sodiumsulphate.Therefore, 15.15g of lead sulphate is precipitated from =142g303g×15.15g=7.1g of sodium sulphate.Therefore, amount of sodium sulphate present is 7.1g.

Answer 2

Answer:

7.1 g of Na2SO4

Explanation:

Reaction : Na2SO4 + Pb(NO3)2 → PbSO4+2NaNO3 (Na = 23, O = 16, S = 32 )

Molar mass of Na2SO4 = 23 x 2 + 32 + 16 x 4 = 142 g

Molar mass of PbSO4 = 207 + 32 + 16 x 4 =303 g

303 g of PbSO4 is precipitated from 142 g of Na2SO4.

∴ 15.15 g of PbSO4 is precipitated from → (142 g ÷ 303 g) x 15.15 g = 7.1 g of Na2SO4.

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