# chemistry

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For these problem, I know how to solve it, but I have some issues with the formula. I know that, I need to write down the formula first, and then I need to balance the equation, then I can solve it. So, can anyone please check my formula? You don't have to solve the problem just please check my formula. Thank you! I know it is too much. But, please help me please! Thank you!

6) How many moles of precipitate are made when 0.03 mol of iron(III) chloride reacts with an excess of silver nitrate?

FeCl3 + AgNO3 -------------> I have no idea how to write this equation

7) How many moles of aluminum metal does it take to react completely with 2.5 mol of nitrogen gas?

Al + N2-------------->AlN2

Cool If 34.5 mol of vanadium(V) chlorate are heated, how many moles of oxygen gas are made?

V+ 5ClO3---------------->V(ClO3)5

9) If excess chromium (VI) oxalate reacts with 3.22 mol of lead (IV) hydoxide, how many mole of new base are made?

Cr2(C2O4)6 + Pb(OH) 4---------->

10) How many moles of sodium phosphate are needed to react with 3.25 mol of lead (IV) nitrate?

Na2PO4 + Pb(NO3)4-------------->

• chemistry -

6) How many moles of precipitate are made when 0.03 mol of iron(III) chloride reacts with an excess of silver nitrate?

FeCl3 + AgNO3 -------------> I have no idea how to write this equation
==> AgCl + Fe(NO3)3

7) How many moles of aluminum metal does it take to react completely with 2.5 mol of nitrogen gas?

Al + N2-------------->AlN2
==> AlN

Cool If 34.5 mol of vanadium(V) chlorate are heated, how many moles of oxygen gas are made?

V+ 5ClO3---------------->V(ClO3)5
V(ClO3)5 ==> VCl5 + O2 (my best guess).

9) If excess chromium (VI) oxalate reacts with 3.22 mol of lead (IV) hydoxide, how many mole of new base are made?
Cr(C2O4)3 + Pb(OH)4 ==> Cr(OH)6 + Pb(C2O4)2

Cr2(C2O4)6 + Pb(OH) 4---------->

10) How many moles of sodium phosphate are needed to react with 3.25 mol of lead (IV) nitrate?

Na2PO4 + Pb(NO3)4-------------->
Na3PO4 + Pb(NO3)4 ==> Pb3(PO4)4 + NaNO3

• chemistry -

I didn't balance any of the equations.

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