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How would you determine yield of the main product if 0.29 g of crcl3-6h2o is used, 1 g of urea is used and 0.8 ml of acac is used. balanced reaction: 2 CrCl3.6H2O + 3 H2NCONH2 + 6 CH3COCH2COCH3 → 2 [cr(ch3cochcoch3)3] + 6 NH4Cl + 12 H2O. actually calculate . show how you determined limiting reagent
Accepted Answer
To determine the yield of the main product, [Cr(CH3COCHCOCH3)3], we need to first identify the limiting reagent. Here's how:
1. Convert masses to moles:
CrCl3.6H2O (0.29 g): (0.29 g) / (266.45 g/mol) = 0.00109 mol
Urea (1 g): (1 g) / (60.06 g/mol) = 0.0167 mol
Acac (0.8 mL): Assuming density of Acac is 1 g/mL, (0.8 mL) / (100.11 g/mol) = 0.0080 mol (Note: Density of Acac can vary, so use its specific density if available)
2. Determine mole ratios based on the balanced equation:
CrCl3.6H2O : [Cr(CH3COCHCOCH3)3] = 2:2 = 1:1
Urea : [Cr(CH3COCHCOCH3)3] = 3:2
Acac : [Cr(CH3COCHCOCH3)3] = 6:2 = 3:1
3. Calculate the theoretical yield of [Cr(CH3COCHCOCH3)3] based on each reactant:
CrCl3.6H2O: (0.00109 mol CrCl3.6H2O)
(1 mol [Cr(CH3COCHCOCH3)3] / 1 mol CrCl3.6H2O) = 0.00109 mol [Cr(CH3COCHCOCH3)3]
Urea: (0.0167 mol Urea)
(2 mol [Cr(CH3COCHCOCH3)3] / 3 mol Urea) = 0.0111 mol [Cr(CH3COCHCOCH3)3]
Acac: (0.0080 mol Acac)
(1 mol [Cr(CH3COCHCOCH3)3] / 3 mol Acac) = 0.0027 mol [Cr(CH3COCHCOCH3)3]
4. Identify the limiting reagent: CrCl3.6H2O produces the least amount of product, making it the limiting reagent.
5. Calculate the theoretical yield of [Cr(CH3COCHCOCH3)3] based on the limiting reagent:
Theoretical Yield = (0.00109 mol [Cr(CH3COCHCOCH3)3])
(400.23 g/mol) = 0.436 g
To determine the actual yield, you would need to carry out the reaction and experimentally measure the mass of the product. The percentage yield is then calculated as (actual yield / theoretical yield)
100%.
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