Balanced Equations, I need a Chemistry Pro help!?
I have a test tomorrow and I dont get this AT ALL! If any one can help me with all of theses or any of these that would be so fantastic thank you!<3
This is the Equation
4FeS₂ + 11O₂ -------> 2Fe₂O₃ + 8SO₂ + 350 KJ
(4FeS₂ (Mwt =120) ) (11O₂ (Mwt =32)) (2Fe₂O₃ (Mwt =160)) (8SO₂(Mwt =64))
1) How much energy in (KJ) will be produced if 15.34 moles of FeS₂ were to completely combust?
2) How many moles of Fe₂ O₃ will be produced if 8.71g of FeS₂ were to react?
3) If 83.5g of FeS₂ were to completely react, how many grams of SO₂ would you theoretically expect to form?
4) If 83.5g of FeS₂ reacted compleatly, as in question #3 and 75.0g of SO₂ were recovered, what would be the % yield of SO₂ ?
5) What is the temperature (in degree Celsius) of a 53.5 g sample of an ideal gas of Mwt = 30.5 g/mol if it is confined in 235 L vessel at a pressure of 0.175 Atm?
6) What will be the new volume when 128 mL of gas at 20.0˚C is heated to 40.0˚C while pressure remains unchanged?
any thing helps thank you!
- DavidLv 68 years agoFavorite Answer
I'm impressed that you can do subscripts here...I don't know how.
1) If 350 kJ are released from 4 moles FeS2, x kJ are released from 15.34 moles FeS2. This is essentially a proportion problem from pre-algebra. Solve it and you get x = 1342 kJ.
2) 8.71 g FeS2 / (120 g/mol) = 0.0726 moles FeS2.
The mole ratio is 4 FeS2 : 2 Fe2O3, so (again, a proportion problem) we get 0.0363 moles Fe2O3
3) 85.3 g FeS2 / (120 g/mol) = 0.711 mole FeS2
The mole ratio is 4 FeS2 : 8 SO2, so (proportionally) we get 1.422 mole SO2
1.422 mole SO2 * (64 g/mol) = 91.0 g SO2
4) 75.0 g / 91.0 g = 82.4% yield. In general, for yield, divide what you actually get by the amount you calculated for complete reaction.
5) Use the ideal gas law, PV=nRT. P = 0.175 atm, V = 235 L, n = 53.5 g / (30.5 g/mol) = 1.75 moles, R = 0.082057 L atm K-1 mol-1.
So T = PV/nR = (0.175 atm) (235 L) / (1.75 mol) (0.082057 L atm K-1 mol-1) = 123 K,
and 123 K - 273.15 = -150°C.
6) Since PV=nRT, we have V1/T1 = nR/P = V2/T2. 20.0°C = 293.15 K and 40.0°C = 313.15 K.
Therefore 128 mL / 298.15 K = x / 313.15 K ---> x = 134 mL.
- 4 years ago
this is Redox reaction C8H18 is impartial. H has +a million fee. then C8 is eighteen- -18/8= -2.25 in CO2 O has fee -2 and C has fee +4 C2.25- - 6.25e = C4+ |6.5 O2 + 4e = 2O2- |4 4C9- - 25e = C16+ |25 O2 + 4e = 2O2- |4 2C8H18 + 25O2 -> 16CO2 + 18H2O
- 8 years ago
1. is a mole to energy proportion...use the coefficients of the reaction 350kJ and 4
2. just plains 3 step stoich...use example 3 in this video
3.same as #2
4. percent yield...just plug in your numbers ...http://www.kentchemistry.com/links/Math/PercentYie...
5.use the ideal gas law...http://www.kentchemistry.com/links/GasLaws/idealGa...
6. use Charles Law...http://www.kentchemistry.com/links/GasLaws/charles...