How many grams of CO2(g) are formed when 1.94 g of C9H16O4 is completely burned in in an oxygen atmosphere

this is the full problem set, its extra credit on our quiz tomorrow and i have no idea how to do it.


How many grams of CO2(g) are formed when 1.94 g of C9H16O4 is completely burned in in an oxygen atmosphere?
mass of CO2 (g)=

How many grams of lead(II) chloride will precipitate when 1.712 g of FeCl3•5H2O is added to 120 mL of a 0.0976 M lead nitrate solution?
mass of PbCl2 (g)=

Hello

find how many moles 1.94 g are: moles = mass/(mass/mole) = 1.94/188.2219 = 0.0103 mole.
= 0.0103 *9 moles C = 0.09276 moles C = 0.09276 moles CO2 = 0.09276*44 g CO2
= 4.08 g <--- ans.
-------
1.712 g = 1.712/252.2048 moles = 0.006788 moles FeCl3•5H2O
120 mL of 0.0976 M = 0.011712 moles Pb(NO3)2
3 Pb(NO3)2 + 2 FeCl3 ---> 2 Fe(NO3)3 + 3 PbCl2

for 0.006788 moles FeCl3 you need 0.006788*3/2 = 0.010182 moles Pb(NO3)2.
You do have 0.011712 moles, so the FeCl3 is limiting.
2 moles FeCl3 give 3 moles PbCl2,
so 0.006788 moles FeCl3 give 0.006788*3/2 moles = 0.010182 moles PbCl2
= 0.010182*278.1164 g PbCl2 = 2.83 g PbCl2 <-- ans.

Regards