Solution:Fe2O3 + 3CaS = Fe2S3 + 3CaOx g 51,9 g(Fe2O3) = 56*2+16*3= 160g/mol(CaO)=40+16=56g/molXmol(CaO)= 51,9/56= 0,96Xmol(Fe2O3)=0,92/3=0,308Xg(Fe2O3)=160*0,308/1=49,28Answer:m(Fe2O3)=49,28g
1 Answers 1 viewsFrom the given info, we know the amount of heat released when 2 moles of Na is reacted. We're asked for the heat per 0.5 moles. So, (368.6 kJ/2mol)0.5000 mol = 92.15...
1 Answers 1 viewsmL of 0.250 M KOH=5.5
1 Answers 1 viewsLitres of O2=72.7L O2
1 Answers 1 viewsMole ratio C2H4:CO2=1:2= 12.5×2=25moles of CO2
1 Answers 1 views4H3PO3 --gt; 3H3PO4 + PH3
1 Answers 1 views__C2H4(g) + __O2(g) → 2CO2(g) + 2H2O(g)Solution:1C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(g)or:C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(g)Answer: 1, 3
1 Answers 1 viewsCuSO4.5H2O + 2KIO3 = Cu(IO3)2 + K2SO4 + 5H2On (KIO3) = 2 x n (CuSO4.5H2O)n = m / MM (CuSO4.5H2O) = 250 g/moln (CuSO4.5H2O) = 0.3500 / 250 = 0.0014...
1 Answers 1 viewsSolution:Balanced chemical equation:C3H8 + 5O2 → 3CO2 + 4H2OAccording to the equation above: n(O2)/5 = n(H2O)/4Thus,n(H2O) = 4 × n(O2) / 5 = (4 × 9.7 mol) / 5 = 7.76 moln(H2O) =...
1 Answers 1 viewsSolution:Balanced chemical equation:2NH3 + H2SO4 → (NH4)2SO4According to the equation above: n[NH3]/2 = n[(NH4)2SO4]Therefore,n[(NH4)2SO4] = n[NH3] / 2 = 30.0 mol / 2 = 15.0 mol (NH4)2SO4Answer: 15.0 moles of ammonium sulfate can...
1 Answers 1 views