A bacterium can convert xylose to lactic acid with a percent yield of $85$ %. What mass of lactic acid would be produced from $50.0 g$ sample of xylose by this bacterium? (At. mass of xylose $(C_{5} H_{10} O_{5}) = 150 g / m o l,$ At. of lactic acid $(C_{3} H_{6} O_{3}) = 90 g / m o l$ ).

58.8 g

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85 g

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50 g

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42.5 g

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Solution

The correct option is D $42.5 g$
Percent Yield is the ratio of actual yield of a substance to the theoretical yield.
Percent Yield $= \frac{Actual Yield}{Theoretical Yield} \times 100$
Given, Theoretical Yield $50 g,$ Percent Yield $= 85$
So, $(\frac{85}{100} \times 50) =$ Actual Yield
$\Rightarrow$ Actual Yield $= 42.5 g$

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A bacterium can convert xylose to lactic acid with a percent yield of 85%. What mass of lactic acid would be produced from 50.0 g sample of xylose by this bacterium? (At. mass of xylose(C5H10O5)=150 g/mol, At. of lactic acid(C3H6O3)=90 g/mol).

The correct option is D 42.5 gPercent Yield is the ratio of actual yield of a substance to the theoretical yield.Percent Yield =Actual YieldTheoretical Yield×100Given, Theoretical Yield 50 g, Percent Yield =85So, (85100×50)= Actual Yield⇒ Actual Yield =42.5 g

A bacterium can convert xylose to lactic acid with a percent yield of $85$ %. What mass of lactic acid would be produced from $50.0 g$ sample of xylose by this bacterium? (At. mass of xylose $(C_{5} H_{10} O_{5}) = 150 g / m o l,$ At. of lactic acid $(C_{3} H_{6} O_{3}) = 90 g / m o l$ ).