How many grams of 3-hydroxypropanal (MW=74) m | vedclass

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(B) The dehydration reaction of $3-$hydroxypropanal is:$HO-CH_2-CH_2-CHO \xrightarrow{\Delta} CH_2=CH-CHO + H_2O$Let the mass of $3-$hydroxypropanal b

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$16$

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(B) The dehydration reaction of $3-$hydroxypropanal is:$HO-CH_2-CH_2-CHO \xrightarrow{\Delta} CH_2=CH-CHO + H_2O$Let the mass of $3-$hydroxypropanal be $x \ g$.Number of moles of $3-$hydroxypropanal $= \frac{x}{74} \ mol$.Theoretical moles of acrolein produced $= \frac{x}{74} \ mol$.Actual moles of acrolein produced $= \frac{7.8 \ g}{56 \ g/mol} \approx 0.1393 \ mol$.Given percentage yield $= 64\% = 0.64$.Actual yield $= 	ext{Theoretical yield} 	imes 	ext{Percentage yield}$.$0.1393 = \left(\frac{x}{74}ight) 	imes 0.64$.$x = \frac{0.1393 	imes 74}{0.64} \approx 16.11 \ g$.Rounding off to the nearest integer,we get $16 \ g$.

How many grams of $3-$hydroxypropanal $(MW=74)$ must be dehydrated to produce $7.8 \ g$ of acrolein $(MW=56)$ $(C_3H_4O)$ if the percentage yield is $64\%$? (Round off to the nearest integer).
[Given: Atomic masses: $C = 12.0 \ u, H = 1.0 \ u, O = 16.0 \ u$]

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How many grams of $3-$hydroxypropanal $(MW=74)$ must be dehydrated to produce $7.8 \ g$ of acrolein $(MW=56)$ $(C_3H_4O)$ if the percentage yield is $64\%$? (Round off to the nearest integer).[Given: Atomic masses: $C = 12.0 \ u, H = 1.0 \ u, O = 16.0 \ u$]