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3 years ago

What is the mass of three moles of chlorine?

Chemistry

2 answers:

Elina [12.6K]3 years ago

4 0

Answer:

The answer to your question is

a) 212.7 g

b) 110 g

Explanation:

a) Mass of three moles of Chlorine

Atomic weight of Chlorine = 35.45 g but chlorine is a diatomic molecule then the atomic weight = 2(35.45) g = 70.9 g

70.9 g of Cl₂ --------------- 1 mol

x --------------- 3 moles

x = 3(70.9)/1

x = 212.7 g

b) mass of one mole of K₂S

Molecular mass = (39 x 2) + (32 x 1)

= 78 + 32

= 110 g

110 g of K₂S ---------------- 1 mol

Genrish500 [490]3 years ago

4 0

The first one I got this

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3 years ago

Honors Stoichiometry Activity Worksheet

vaieri [72.5K]

Explanation: Here, we will be considering 1 cup is equal to 1 mol.

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Moles of sugar present in 196.86 g of water $=\frac{196.86 g}{225 g/mol}=0.8749 mol$

Moles of lemon juice present in 193.37 g of water $=\frac{193.37 g}{257.83 g/mol}=0.7499 mol$

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As we can see that number of moles of lemon juice are limited.

So, we will consider the reaction will complete in accordance with moles of lemon juice.

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$\frac{2}{1}\times 0.7499 mol = 1.4998 mol$ of water

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Water remained unused = 946.36 g - 354.8376 g =591.5223 g

1 mole lemon juice reacts with mol of sugar,then 0.7499 mol of lemon juice will react with:

$\frac{1}{1}\times 0.7499 mol = 0.7499 mol$ of water

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Sugar remained unused = 196.86 g - 28.1325 g

1 mole of lemon juice gives 4 moles of lemonade.

Then 0.7499 mol of lemon juice will give:

$\frac{4}{1}\times 0.7499 mol=2.996 mol$ of lemonade

Mass of lemonade obtained = 2.996 mol × 719.42 g/mol = 2157.9722 g

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Experimental yield of lemonade = 2050.25 g

Percentage yield of lemonade:

$\frac{\text{Experimental yield}}{\text{theoretical yield}}\times 100$

$\frac{2050.25 g}{2157.9722 g}\times 100=95.00\%$

The percentage yield of lemonade is 95% and ingredient which remained unused were water and sugar.

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Soloha48 [4]

Answer:

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