Answer:
72.53% is the yield of CrCl3
Explanation:
Given
Reaction:
Cr2O3(s) + 3 CCl4(l) → 2 CrCl3(s) + 3 COCl2(aq)
CCl4 is in excess and 17.6g Cr2O3 present
The reaction yields 26.6g of CrCl3
To Find:
% yields of the reaction
Also given
Molar mass of CrCl3 = 158.35g/mol
Molar mass of Cr2O3 = 152.00 g/mol
By the stoichiometry of the reaction
1 mole of Cr2O3 gives 2 moles of CrCl3
0r
1 x1 52 g of Cr2O3 gives 2x 158.35 g of CrCl3
= 1 52 g of Cr2O3 gives 316.70 g of CrCl3
17.6 g of Cr2O3 gives (17.6÷152) × 316.70 g CrCl3
= 36.67 g CrCl3
but actual yield is only 26.6g
so % yield is (26.6 ÷÷ 36.67) × 100
= 72.53% is the yield of CrCl3
The molarity of KOH is 0.1055 M
<u><em> calculation</em></u>
Step 1: write the equation for reaction between H₂C₂O₄.2H₂O and KOH
H₂C₂O₄.2H₂O + 2 KOH → K₂C₂O₄ +4 H₂O
step 2: find the moles of H₂C₂O₄.2H₂O
moles = mass÷ molar mass
from periodic table the molar mass H₂C₂O₄.2H₂O= (1 x2) +(12 x2) +(16 x4) + 2(18)=126 g/mol
= 0.2000 g ÷ 126 g/mol =0.00159 moles
step 3: use the mole ratio to calculate the moles of KOH
H₂C₂O₄.2H₂O : KOH is 1:2
therefore the moles of KOH =0.00159 x 2 = 0.00318 moles
step 4: find molarity of KOH
molarity = moles/volume in liters
volume in liters = 30.12/1000=0.03012 L
molarity is therefore = 0.00318/0.03012 =0.1055 M
Answers are:
1) shift to fossil fuels or coals with lower amount of sulfur.
2) removal (s<span>pray dry system</span>) of the sulfur oxide after combustion, removal sulfur during combustion (fluidised bed combustion) or reducing the sulpfur before combustion (washing the coal)<span>.
3) </span><span>improve efficiency of conversion of fuel and coal to electric power.
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Answer:
umm...what is the question though?
Explanation:
