A student performs the reduction of 4-nitrobenzaldehyde (151.12 g/mol) with sodium borohydride (37.83 g/mol) in the presence of ethanol. The student mixes 4.13 g of 4-nitrobenzaldehyde with 0.700 g of sodium borohydride and obtains 1.85 g of 4-nitrobenzyl alcohol (153.14 g/mol). What is the percent yield of this reaction

65.4%

Explanation:

The redox reaction is a 1:1:1 reaction because the reagents suffer a double displacement reaction, and the substance that is substituted have the same charge (H + and Br-), thus, we first need to know which of the reagents is the limiting.

Let's test the 4-nitrobenzaldehyde as the limiting. The mass needed for sodium borohydride (m) is the mass given of 4-nitrobenzaldehyde multiplied by the stoichiometric mass of sodium borohydride divided by the stoichiometric mass of 4-nitrobenzaldehyde. The stoichiometric mass is the number of moles in the stoichiometric representation (1:1:1) multiplied by the molar mass, so:

m = (4.13 * 37.83*1) / (151.12*1)

m = 1.034 g

So, the mass needed of the other reagent is larger than the mass that was given, so, it will be the limiting, and the stoichiometric calculus must be done with it.

The mass of the product that was expected is then:

m = (0.700*153.14*1) / (37.83*1)

m = 2.83 g

The percent yield is the mass that was formed divided by the expected mass, and then multiplied by 100%:

%yield = (1.85/2.83) * 100%

%yield = 65.4%