Use the following reaction: C4H9OH NaBr H2SO4 C4H9Br NaHSO4 H2O If 15.0 g of C4H9OH react with 22.4 g of NaBr and 32.7 g of H2SO4 to yield 17.1 g of C4H9Br, what is the percent yield of this reaction

Percent yield = 61.58%

Explanation:

C4H9OH + NaBr + H2SO4 --> C4H9Br + NaHSO4 + H2O

From the reaction;

1  mol of C4H9OH reacts with 1 mol of NaBr and 1 mol of H2SO4 to form 1 mol of  C4H9Br

We have to obtain the limiting reagent. We do this by calculatig the mols of each reactant in the reaction;

Number of moles = Mass / Molar mass

C4H9OH; 15 / 74g/mol = 0.2027

NaBr; 22.4 / 102.89 = 0.2177

H2SO4; 32.7 / 98 = 0.3336

In this case, the limiting reagent is C4H9OH, it determines the amount of product formed.

From the stoichometry of the reaction; one mole of C4H9OH forms one mole of C4H9Br. This means 0.2027 mol of C4H9Br was formed.

Mass = Number of moles * Molar mass

Mass of C4H9Br formed = 0.2027 * 137

Mass = 27.77g

Percent yield =  (Practical yield /  theoretical yeield ) * 100%

Percent Yield = 17.1 / 27.77 * 100

Percent yield = 61.58%

explanation:

metals have excess of electrons whereas non-metals have deficiency of electrons. thus, in order to gain stability metals easily donate their electrons to the non-metals.

for example, electronic configuration of sodium is and electronic configuration of phosphorous is .

when a phosphorous atom and sodium atom combine together then sodium will donate is extra one electron to the phosphorous atom. as a result there will be formation of ionic bond between sodium and phosphorous.

the reaction will be as follows.

there, are three sodium sodium atoms and each of them will donate its one extra electron to phosphorous atom. thus, the sub-shell of phosphorous gets completely filled.  

thus, we can conclude that electrons are transferred from atoms of sodium to atoms of phosphorous. this transfer makes the sodium atoms donate one electron and the phosphorous atoms to gain one electron. as a result, the sodium and phosphorus atoms strongly attract each other.

nuclear fission produces elements that are heavier than helium.

the elements that are used in nuclear fission and their products are much heavier than helium.