Question

Honors Stoichiometry Activity Worksheet

Answer 1

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

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$=\frac{193.37 g}{257.83 g/mol}=0.7499 mol$

Moles of lemonade in 2050.25 g of water$=\frac{2050.25 g}{719.42 g/mol}=2.8498 mol$

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

Mass of water used =  1.4998 mol × 236.59 g/mol=354.8376 g

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.