1) since we are given percentages, we can assume we have 100 grams of the molecule.
55.6 % Cu ----> 55.6 grams Cu
16.4 % Fe------> 16.4 grams Fe
28.0% S--------> 28.0 grams S
2) convert each gram to moles using the molar masses given



3) we divide the smallest value of moles (0.293) to each one.
Cu --> 0.876 / 0.293= 3
Fe---> 0.293 / 0.293= 1
S-----> 0.875 / 0.293= 3
4) let's write the empirical formula
Cu₃FeS₃
2)
formula equation: Pb(NO₃)₂(aq) + 2KI(aq) ----> PbI₂(s) + 2KNO₃(aq)
total ionic equation: Pb⁺² + 2NO₃⁻¹ + 2K⁺¹ + 2I⁻¹----> PbI₂(s) + 2K⁺¹ + 2NO₃⁻¹
net ionic equation: Pb⁺² + 2 I⁻¹----> PbI₂(s)
3)
formula equation: Zn(NO₃)₂(aq) + K₂CO₃(aq) -----> ZnCO₃(s) + 2KNO₃(aq)
total ionic equation: Zn⁺² + 2NO₃⁻¹ + 2K⁺¹ + CO₃⁻² ---> ZnCO₃ (s) + 2K⁺¹ + 2NO₃⁻¹
net ionic equation: Zn⁺² + CO₃⁻² ----> ZnCO₃ (s)
note: if I did not specify the state of the molecule in the reaction, you can assume they are aqueous unless state otherwise.
To solve this we use the equation,
M1V1 = M2V2
where M1 is the concentration of the stock solution, V1 is the volume of the stock solution, M2 is the concentration of the new solution and V2 is its volume.
.675 M x V1 = .25 M x 1.3 L
V1 = 0.48 L or 480 mL