Question

For each of the reactions, calculate the mass (in grams) of the product formed when 15.82 g of the underlined reactant completely reacts. Assume that there is more than enough of the other reactant.

Express your answer using four significant figures.

1) 2K(s)+Cl2(g)−−−−−→2KCl(s)

2) 2K(s)+Br2(l)−−−−−→2KBr(s)

3) 4Cr(s)+3O2(g)−−−−−→2Cr2O3(s)

Answer 1

1) 2K(s)+Cl2(g)−−−−−→2KCl(s)

from this reaction ( moles of Cl2 = 15.82/71 ==> 0.223 moles)

1 mole of Cl2 gives -----> 2 moles of KCl

0.223 moles of Cl2 gives ----> (2 X 0.223)/1 = 0.446 moles

mass of KCl = moles X molar mass = 0.446 X 74.5 = 16.8 grams.

2) 2K(s)+Br2(l)−−−−−→2KBr(s)

from this reaction ( moles of Br2 = 15.82/159.8 ==> 0.099 moles)

1 mole of Br2 gives -----> 2 moles of KBr

0.099 moles of Br2 gives ----> (2 X 0.099)/1 = 0.198 moles

mass of KBr = moles X molar mass = 0.198 X 119 = 23.6grams.

3)4Cr(s)+3O2(g)−−−−−→2Cr2O3(s)

from this reaction ( moles of O2 = 15.82/32 ==> 0.49 moles)

3 moles of O2 gives -----> 2 moles of Cr2O3

0.49 moles of O2 gives ----> (2 X 0.49)/3 = 0.327 moles

mass of Cr2O3 = moles X molar mass = 0.327 X 151.9 = 49.7 grams.