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elixir [45]
3 years ago
13

During a chemical reaction, 25.0 g of propane burns with 90.7 g of oxygen to form carbon dioxide and water. what mass of carbon

dioxide and water is formed? during a chemical reaction, 25.0 g of propane burns with 90.7 g of oxygen to form carbon dioxide and water. what mass of carbon dioxide and water is formed? 115.7 g 90.7 g 65.7 g 25.0 g
Chemistry
1 answer:
dusya [7]3 years ago
5 0

The chemical equation for the combustion of propane can be represented as,

C_{3}H_{8}(g) + 5O_{2}(g)-->3CO_{2}(g) + 4H_{2}O(g)

1 mol propane requires 5 mol oxygen for complete reaction to produce 3 mol carbon dioxide and 4 mol water.

Mass of propane = 25.0 g

Mass of oxygen = 90.7 g

The total mass of reactants = 25.0 g + 90.7 g = 115.7 g

According to Law of conservation of mass, the total mass of reactants will be equal to the total mass of products formed in a chemical reaction.

Therefore, here the mass of water and carbon dioxide will be 115.7 g

Correct answer: a. 115.7 g



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Honors Stoichiometry Activity Worksheet
vaieri [72.5K]

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.

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