Can someone please help me solve these 2 problems?

Question

They're stoichiometry problems & I tried figuring it out but I can't seem to get the correct answers.
Problem1 : If 662g of Pb(NO3)2 undergo this reaction, Pb(NO3)2+Na3(PO4)2+Na(NO3),how many grams of Na(NO3) are produced?
2nd problem: If 530g of Ag(NO3) undergo this reaction, Ag(NO3)+feCl3-->AgCl+Fe(NO3)3, how many grams of Fe(NO3)3 are produced?
They have to have also, the conversion factors & be balanced. All help would be appreciated.

Answer 1

3Pb(NO3)2 +2 Na3PO4 >>6NaNO3 + Pb3(PO4)2

MM(Pb(NO3)2=331 g/mol

662/331= 2 moles

the ratio between Pb(NO3)2 and NaNO3 is 3 : 6

3 : 6 = 2 : x

x = 4 moles

MM(NaNO3)=85 g/mol

85(4)=340 g



3AgNO3 + FeCl3 >> 3AgCl + Fe(NO3)3

MM AgNO3 = 170 g/mol

530/170 = 3 moles AgNO3

so 1 mole Fe(NO3)3 will be produced

MM Fe(NO3)3 = 241.8 g/mol

We will have 241.8 g

Answer 2

I'm assuming the first one is lead nitrate plus sodium phosphate going to lead phosphate and sodium nitrate.

3 Pb(NO3)2 + 2 Na3PO4 ---> Pb3(PO4)2 + 6 NaNO3

662 g Pb(NO3)2 * 1 mol/331.2 g * 2 moles NaNO3/1 mol Pb(NO3)2 * 85.0 g/mol = 340 g NaNO3

3 AgNO3 + FeCl3 ---> 3 AgCl + Fe(NO3)3

530 g AgNO3 * 1 mol/169.9 g * 1 mol Fe(NO3)3/3 mole AgNO3 * 241.8 g/mol = 251 g Fe(NO3)3

Answer 3

for problem 1 the equation seems to be wrong as no products are given in the equation.

for problem 2, the balanced equation is

3Ag(NO3) + FeCl3 --> 3AgCl + Fe(NO3)3

no of mol of Ag(NO3)3 = 530/ (108.0 + 14.0 + 3*16.0)
= 530 /170.0
= 3.117647
since Ag(NO3) = Fe(NO3)3

no of mol of Fe(NO3)3 = 3.117647
mass of Fe(NO3)3 produced = 3.117647 * (55.8+ 3* 62.0)
= 753.847g

if possible pls include the ans so that we can check our answers too.