For each of the reactions, calculate the mass (in grams) of the product formed when 15.67 g of the underlined reactant completely reacts. Assume that there is more than enough of the other reactant.2K(s)+Cl2(g)−−−−−→2KCl(s)2K(s)+Br2(l)−−−−−→2KBr(s)4Cr(s)+3O2(g)−−−−−→2Cr2O3(s)2Sr(s)−−−−+O2(g)→2SrO(s)

Question
Asked Oct 7, 2019

For each of the reactions, calculate the mass (in grams) of the product formed when 15.67 g of the underlined reactant completely reacts. Assume that there is more than enough of the other reactant.

2K(s)+Cl2(g)−−−−−→2KCl(s)

2K(s)+Br2(l)−−−−−→2KBr(s)

4Cr(s)+3O2(g)−−−−−→2Cr2O3(s)

2Sr(s)−−−−+O2(g)→2SrO(s)

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Step 1

(1) In the given reaction, the underlined reactant is Cl2 (g).

Amount of chlorine is 15.67 g.

Now, calculate the number of moles of chlorine as follows:

 

mass of Cl
Moles of Cl2 molar mass of Cl2
15.67 g
70.906 g-mol1
0.22 mol
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mass of Cl Moles of Cl2 molar mass of Cl2 15.67 g 70.906 g-mol1 0.22 mol

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Step 2

15.67 g Cl2 gives 0.22 moles.

1 mole of Cl2 gives 2 moles of KCl.

So, 0.22 moles of Cl2will give 0.44 moles KCl.

Therefore, the mass of 0.44 moles of KCl can be calculated as,

Thus, the mass of KCl is 32.80 g.

Mass of KCl
moles of KCl x molar mass of Kci
= 0.44 mol x 74.55 g-mol-1
= 32.80 g
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Mass of KCl moles of KCl x molar mass of Kci = 0.44 mol x 74.55 g-mol-1 = 32.80 g

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Step 3

(2) The underlined reactant is Br2 (l).

Amount of bromine is 15.67 g.

Now, calcul...

mass of bromine
Moles of bromine = molar mass ofbromine
15.67 g
159.808 g-mol1
=0.098 mol
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mass of bromine Moles of bromine = molar mass ofbromine 15.67 g 159.808 g-mol1 =0.098 mol

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