Answer to Question #139294 in General Chemistry for jx

Question

two pounds of 44.4% potassium bromate solution at 70 degrees celsius has been cooled at 10 degrees celsius. Estimate the amount of crystals to be recovered

Answer 1

Solubility of KBr at 10$C^\circ$ is 60 g of KBr /100 g water

mass percent of KBr in this solution is

$w(KBr) = \frac{mass (KBr)}{mass (solution)}=\frac{60}{60+100}=0.375$

1) Find mass of KBr in the given solution

$2 pounds (solution)\times \frac{453.592 g}{1 pound}= 907.185 g (solution)$

$mass(KBr) = 907.185 g \times 0.444 =402.790 g$

2) x- mass of the crystalls

mass percent of KBr in the solution after the crystals precipitated

$mass \% = \frac{mass(KBr)}{mass(solution)}=\frac{402.790-x}{907.185-x}$

what is equal to mass percent of KBr at 10 $C^\circ$ $, 0.375$

solve the equation

$0.375=\frac{402.790-x}{907.185-x}$

x$=100.16$

the mass of the crystals is 100.16 g