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attashe74 [19]
3 years ago
8

The predicted yield for a reaction was 10g of product. If the percentage yield for this reaction was 50%, how much product was f

ormed?​
Chemistry
1 answer:
Pachacha [2.7K]3 years ago
7 0

The quantity of the product that was formed was 5g

From the question,

We are to determine the quantity of product formed. That is, the actual yield.

Using the formula

Percentage \ yield =\frac{Actual \ yield}{Theoretical\ yield} \times 100\%

From the given information

Theoretical yield = 10g

Percentage yield = 50%

Putting the parameters into the formula, we get

50\% = \frac{Actual\ yield}{10} \times 100\%

Then,

50\% = Actual\ yield \times 10\%

∴ Actual yield = \frac{50\%}{10\%}

Actual yield = 5g

Hence, the quantity of the product that was formed was 5g

Learn more here: brainly.com/question/17127987

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A solution is prepared by mixing 525 mL of ethanol with 597 mL of water. The molarity of ethanol in the resulting solution is 8.
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Answer:

\Delta V = 234.736\,mL

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n_{C_{2}H_{5}OH} = 4.985\,mol

The mass and volume of ethanol in the solution are, respectively:

m_{C_{2}H_{5}OH} = (4.985\,mol)\cdot \left(46.07\,\frac{g}{mol} \right)

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V_{C_{2}H_{5}OH} = \frac{229.658\,g}{0.7893\,\frac{g}{mL} }

V_{C_{2}H_{5}OH} = 290.964\,mL

The difference between the total volume of water and ethanol mixed to prepare the solution and the actual volume of solution is:

\Delta V = (525\,mL+597\,mL) - (597\,mL + 290.964\,mL)

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