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2 years ago

If 15.6 g of hydrate are heated and only 11.7 g of anhydrous salt remain, calculate the % of water lost.

Chemistry

1 answer:

Elodia [21]2 years ago

8 0

Answer:

Percent loss of water = 25%

Explanation:

Given data:

Mass of hydrated salt = 15.6 g

Mass of anhydrous salt = 11.7 g

Percentage of water lost = ?

Solution:

First of all we will calculate the mass of water in hydrated salt.

Mass of water = Mass of hydrated salt - Mass of anhydrous salt

Mass of water = 15.6 g - 11.7 g

Mass of water = 3.9 g

Now we will calculate the percentage.

Percent loss of water = mass of water / total mass × 100

Percent loss of water = 3.9 g/ 15.6 g × 100

Percent loss of water = 25%

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100 POINTS PLEASE HELP!! Honors Stoichiometry Activity Worksheet Instructions: In this laboratory activity, you will taste test

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Answer:

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=

236.59g/mol

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=4mol

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

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=2.8498mol

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

×0.7499mol=1.4998mol of water

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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

×0.7499mol=0.7499mol 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

×0.7499mol=2.996mol 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

theoretical yield

Experimental yield

×100

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

2157.9722g

2050.25g

×100=95.00%

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Answer:

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