the oxidation of zinc sulfide produces the products zinc oxide and sulfur dioxide, thus the equation below
ZnS + O2 → ZnO + SO2
Since zinc and sulfur are balanced multiply the coefficients of compounds containing these elements by 2 and proceed to balance oxygen
2ZnS + 3O2 → 2ZnO + 2SO2
From the balanced equation we see that 3 moles of oxygen gas are needed to oxidize 2 moles of zinc sulfide completely.
Molar mass of zinc sulfide = 65.4 + 32.1 = 97.5
Moles of zinc sulfide = 380/97.5 = 3.90
Molar mass of O2 = 2(16) = 32
Number of moles of oxygen gas = 176/32 = 5.5
Number of moles of oxygen gas needed to oxidize 3.90 moles of zinc sulfide = 3(3.90)/2 = 5.85 > 5.5
so oxygen is the limiting reagent.
5.5 moles of oxygen can oxidize 2(5.5)/3 = 11/3 moles of zinc sulfide.
Mass of zinc sulfide oxidized = (11/3)(97.5) = 357.5g
Thus 380.0g - 375.5g = 4.5g of zinc sulfide remains.
For the second reaction note the oxidation states of cobalt (+3), sulfide(-2) and ammonium (+1) ions.
Co(NO3)3(aq) + (NH4)2S (aq)→ Co2S3(s) + NH4NO3(aq)
2Co(NO3)3(aq) + (NH4)2S(aq) → Co2S3(s) + NH4NO3(aq) (balance cobalt)
2Co(NO3)3(aq) +3 (NH4)2S(aq) → Co2S3(s) + NH4NO3(aq) (balance sulfur)
2Co(NO3)3(aq) + 3(NH4)2S(aq) → Co2S3(s) + 6 NH4NO3(aq) (balance nitrogen and note hydrogen and oxygen are balanced)
The last equation is the decomposition of dinitrogen hydroxide to form nitrogen oxide and oxygen gas.
N2O5 → NO2 + O2
Balance nitrogen first then oxygen
N2O5 → 2NO2 + O2
2N2O5 → 4NO2 + O2