<span>4FeS2 + 11O2 = 2Fe2O3 + 8SO2</span>
Percent yield is calculated as the actual yield divided by the theoretical yield multiplied by 100.
Actual yield = 55 g ( 1 mol / 159.69 g ) = 0.34 mol Fe2O3
To find for the theoretical yield, we first determine the limiting reactant.
100 g O2 ( 1 mol / 32 g) = 3.13 mol O2
200 g FeS2 (1 mol / 119.98g) = 1.67 mol FeS2
Therefore, the limiting reactant is O2.
Theoretical yield = 3.13 mol O2 ( 2 mol Fe2O3 / 11 mol O2 ) = 0.57 mol Fe2O3
Percent yield = (0.34 mol / 0.57 mol) x 100 = 59.74%
Answer:
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Explanation:
Use the given functions to set up and simplify 173 ° C .
1.5 =
CH4 = CH4
4.4 = CH4
173 ° C = CH4
The molarity of aqueous lithium bromide, LiBr solution is 0.2 M
We'll begin by calculating the number of mole of Pb(NO₃)₂ in the solution.
- Volume = 10 mL = 10 / 1000 = 0.01 L
- Molarity of Pb(NO₃)₂ = 0.250 M
- Mole of Pb(NO₃)₂ =?
Mole = Molarity x Volume
Mole of Pb(NO₃)₂ = 0.25 × 0.01
Mole of Pb(NO₃)₂ = 0.0025 mole
Next, we shall determine the mole of LiBr required to react with 0.0025 mole of Pb(NO₃)₂
Pb(NO₃)₂ + 2LiBr —> PbBr₂ + 2LiNO₃
From the balanced equation above,
1 mole of Pb(NO₃)₂ reacted with 2 mole of LiBr.
Therefore,
0.0025 mole of Pb(NO₃)₂ will react with = 2 × 0.0025 = 0.005 mole of LiBr
Finally, we shall determine the molarity of the LiBr solution
- Mole = 0.005 mole
- Volume = 25 mL = 25 / 1000 = 0.025 L
- Molarity of LiBr =?
Molarity = mole / Volume
Molarity of LiBr = 0.005 / 0.025
Molarity of LiBr = 0.2 M
Learn more about molarity: brainly.com/question/10103895
The answer to the question is- Fe2O3