Aqueous hydrobromic acid HBr reacts with solid sodium hydroxide NaOH to produce aqueous sodium bromide NaBr and liquid water H2O . If 0.555g of sodium bromide is produced from the reaction of 2.4g of hydrobromic acid and 0.77g of sodium hydroxide, calculate the percent yield of sodium bromide. Be sure your answer has the correct number of significant digits in it.

28%

Explanation:

We must obtain the balanced reaction equation as this is the basis for all the calculations that follows. From the balanced reaction equation, 1 mole of hydrobromic acid reacts with one mole of sodium hydroxide to yield one mole each of sodium bromide and water.

Using simple proportion, we obtain the mass of sodium bromide produced by sodium hydroxide since it is the limiting reactant. This is now used to calculate the actual yield of sodium bromide and subsequently the percentage yield as shown.

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.