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3 years ago

What is the empirical formula of the compound that is 25.3% magnesium and 74.7% chlorine

Chemistry

1 answer:

Margarita [4]3 years ago

4 0

Answer: $MgCl_2$

Explanation:

25.3% Mg

74.7% Cl

First step: change % to g

25.3g Mg

74.7g Cl

Second step: calculate g/mol of each compound. You can do this by using the atomic mass.

$25.3gMg(\frac{1mol}{24.30g})=1.04mol$

$74.7gCl(\frac{1mol}{35.45g} )=2.11mol$

Third step: determine the lowest number and divide everything by it. Of the result, extract whole number only.

$Mg=\frac{1.04}{1.04} =1$

$Cl=\frac{2.11}{1.04}=2$

Fourth step: Write each compound with their respective number below.

This empirical formula should be: $MgCl_2$

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For the reaction: 3 KClO3 to 3 KCL + 4O2, How many miles of potassium chloride are produced from 9.3 grams of potassium chloride

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<u>Explanation:</u>

First we have to write the balanced equation as,

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$Moles of $K C l O_{3}=9.3 g$ of $K C l O_{3} \times \frac{1 \text { mol of } K C l O_{3}}{122.55 g K C l O_{3}}=0.076$ moles$

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From the moles of KCl, we have to find the mass using its molar mass by multiplying them as,

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