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4 years ago

Consider the following reaction:

Chemistry

1 answer:

eimsori [14]4 years ago

8 0

Answer:

1) a. 2Fe(s) + 3Cl₂(g) → 2FeCl₃(s).

2) a. 0.4477 mol.

b. 0.4477 mol.

c. 72.61 g.

Explanation:

1) Consider the following reaction: iron (s) + chlorine (g) à iron (III) chloride

a. Write the balanced chemical equation.

The balanced equation should apply the law of conversation of mass that the no. of different atoms is equal in both sides of the reaction (reactants and products sides).

So, the balanced chemical equation is:

2Fe(s) + 3Cl₂(g) → 2FeCl₃(s).

It is clear that 2 mol of Fe(s) react with 3 mol of Cl₂(g) to produce 2 mol of FeCl₃(s).

2) 25.0 g of iron reacts with excess chlorine gas.

a. Calculate the moles of iron reactant.

The no. of moles of Fe (n) can be calculated using the relation:

n = mass/molar mass = (25.0 g)/(55.845 g/mol) = 0.4477 mol.

b. Calculate the moles of iron (III) chloride.

Using cross multiplication:

2 mol of Fe produce → 2 mol of FeCl₃, from stichiometry.

∴ 0.4477 mol of Fe produce → 0.4477 mol of FeCl₃.

∴ The no. of moles of iron (III) chloride produced is 0.4477 mol.

c. Calculate the mass of iron (III) chloride.

We can calculate the mass of iron (III) chloride produced using the relation:

mass of iron (III) chloride = (no. of moles)*(molar mass) = (0.4477 mol)*(162.2 g/mol) = 72.61 g.

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2 water + sugar + lemon juice → 4 lemonade

Moles of water present in 946.36 g of water $=\frac{946.36 g}{236.59 g/mol}=4 mol$

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.

1 mole lemon juice reacts with 2 mol of water,then 0.7499 mol of lemon juice will react with:

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

Mass of sugar used = 0.7499 mol × 225 g/mol = 168.7275 g

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

Theoretical yield of lemonade = 2157.9722 g

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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