________ grams of 3-Hydroxy propanal (MW = 74) must be dehydrated to produce 7.8 g of acrolein (MW = 56)(C$$_3$$H$$_4$$O) if the percentage yield is 64. (Round off to the Nearest Integer).
[Given: Atomic masses: C: 12.0u, H: 1.0u, O: 16.0u]

The dehydration of 3-hydroxypropanal produces acrolein (propenal) and water: $$\text{HOCH}_2\text{CH}_2\text{CHO} \xrightarrow{-\text{H}_2\text{O}} \text{CH}_2{=}\text{CHCHO}$$. We are given that the molecular weight of 3-hydroxypropanal is 74 g/mol and that of acrolein is 56 g/mol.

First, we find the moles of acrolein produced: $$n_{\text{acrolein}} = \frac{7.8}{56} = 0.13929 \text{ mol}$$.

Since the reaction is a 1:1 molar conversion, the theoretical moles of 3-hydroxypropanal needed (if yield were 100%) would also be 0.13929 mol. However, the percentage yield is only 64%, which means only 64% of the reactant that could have converted actually did produce the desired product. To get the required amount of product at 64% yield, we need more reactant: $$n_{\text{reactant}} = \frac{0.13929}{0.64} = 0.21764 \text{ mol}$$.

Converting to grams: $$\text{mass} = 0.21764 \times 74 = 16.1 \text{ g}$$.

Rounding to the nearest integer, the answer is $$16$$ grams.