Question

1) For the reaction, calculate how many grams of the product form when 24.4g of Br2 completely reacts.
Assume that there is more than enough of the other reactant.


2K(s)+Br2(l)?2KBr(s)

2)For the reaction, calculate how many grams of the product form when 24.4g of O2 completely reacts.
Assume that there is more than enough of the other reactant.
4Cr(s)+3O2(g)?2Cr2O3(s)

Answer 1

Given the reaction 2K(s) + Br₂(l) -----> 2KBr(s)

Also given to find out the amount of produc formed from 24.4g of bromine molecule

from the reaction we can say that one mole of bromine reacts with 2 moles of potassium to give 2 moles of potassium bromide...

that is 159.8080g of bromine molecule gives 238.004g of potassium bromide

then 24.4g of bromine molecule gives how much of potassium bromide..

159.8080 ----> 238.004

24.4 ------>? of potassium bromide

24.4 x 238.004 /159.8080 = 36.3392g of potassium bromide...

therefore, 24.4g of bromine molecule produces 36.3392g of potassium bromide...

Also given the reaction 4Cr(s) + 3O₂(g) -----> 2Cr₂O₃

given to find out the amount of product formed form 24.4g of O₂ molecule...

From the reaction we can say that 3 moles of oxygen reacts with 4 moles of chromium to give 2 moles of chromium oxide...

one mole of oxygen is of 16g..then 3 moles of oxygen molecule is equal to 48g..

one mole of chromium oxide is of 151.99g... then 2 moles of chromium oxide is equal to 303.98g..

thus... 48g of oxygen molecule gives 303.98g of chromium oxide

then 24.4g of oxygen moleucle should give how much of chromium oxide..

48g ---> 303.98

24.4g ----->?

303.98 x 24.4/48 = 154.52g...

therefore, 154.52g of chromium oxide is produced from 24.4g of oxygen.....