In the following reaction, how many grams of lead(II) nitrate Pb(NO3), will produce 425 grams of sodium nitrate (NaNO3)?
Pb(NO3)2(aq) + 2 NaBr(aq) - PbBr2(s) +2 NaNO3(aq)
The molar mass of Pb(NO3)2 is 331.21 grams and that of NaNO3 is 85 grams.

Answer is: (A) 828.03 grams of lead(II) nitrate.

Balanced chemical reaction:

Pb(NO₃)₂(aq) + 2NaBr(aq) → PbBr₂(s) + 2NaNO₃(aq).

m(NaNO₃) = 425 g; mass of sodium nitrate.

M(NaNO₃) = 85 g/mol; molar mass of sodium nitrate.

n(NaNO₃) = m(NaNO₃) ÷ M(NaNO₃).

n(NaNO₃) = 425 g ÷ 85 g/mol.

n(NaNO₃) = 5 mol; amount of sodium nitrate.

From balanced chemical reaction: n(NaNO₃) : n(Pb(NO₃)₂) = 2 : 1.

n(Pb(NO₃)₂) = 5 mol ÷ 2.

n(Pb(NO₃)₂) = 2,5 mol; amount of lead(II) nitrate.

m(Pb(NO₃)₂) = n(Pb(NO₃)₂) · M(Pb(NO₃)₂)

m(Pb(NO₃)₂) = 2.5 mol · 331.21 g/mol.

m(Pb(NO₃)₂) = 828.03 g.

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i believe you are right for all except the 3rd ( look at sqdancefan his answer is much better)