Saved Bookmarks
| 1. |
Potassium bromide, KBr containss, 32.9% by mass of potassium. If 6.40 g of bromine reacts with 3.60 g of potassium, calculate the number of moles of potassium which combine with bromine to form KBr. |
|
Answer» SOLUTION :In KBr, 32.9 g of K are combined with 67.1 g of Br `THEREFORE"3.6 g of K will combine with "Br_(2)=(67.1)/(32.9)xx3.6g=7.34" g which is not PRESENT"` `"or6.4 g or "Br_(2)" will combine with K"=(32.9)/(67.1)xx6.4=3.14g` Thus, `Br_(2)` is the limiting reactant. `"K reacted= 3.14 g"=(3.14)/(36)" MOLE = 0.08 mole"` (K left unreacted `=(3.60-3.14)g=0.46g`) |
|