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3 years ago

Balance the following equation mg(s)+ HCl H2)g)+MgCl2

Chemistry

1 answer:

Marat540 [252]3 years ago

6 0

<h3>Given equation:-</h3>

$\\ \sf\longmapsto Mg+HCl\longrightarrow MgCl_2+H_2$

<h3>BALANCED EQUATION:-</h3>

$\\ \sf\longmapsto Mg+2HCl\longrightarrow MgCl_2+H_2$

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Hydroxylamine hydrochloride is a powerful reducing agent which is used as a polymerization catalyst. It contains 5.80 mass % H,

ludmilkaskok [199]

<u>Answer:</u> The empirical formula for the given compound is $H_{4}O_1N_1Cl_1=H_4NOCl$

<u>Explanation:</u>

We are given:

Percentage of H = 5.80 %

Percentage of O = 23.02 %

Percentage of N = 20.16 %

Percentage of Cl = 51.02 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of H = 5.80 g

Mass of O = 23.02 g

Mass of N = 20.16 g

Mass of Cl = 51.02 g

To formulate the empirical formula, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{5.80g}{1g/mole}=5.80moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{23.02g}{16g/mole}=1.44moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{20.16g}{14g/mole}=1.44moles$

Moles of Chlorine = $\frac{\text{Given mass of Chlorine}}{\text{Molar mass of Chlorine}}=\frac{51.02g}{35.5g/mole}=1.44moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.44 moles.

For Hydrogen = $\frac{5.80}{1.44}=4.03\approx 4$

For Oxygen = $\frac{1.44}{1.44}=1$

For Nitrogen = $\frac{1.44}{1.44}=1$

For Chlorine = $\frac{1.44}{1.44}=1$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of H : O : N : Cl = 4 : 1 : 1 : 1

Hence, the empirical formula for the given compound is $H_{4}O_1N_1Cl_1=H_4NOCl$

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<h3>Molar mass of Potassium Nitrate:-</h3>

$\\ \large\sf\longmapsto KNO_3$

$\\ \large\sf\longmapsto 39u+14u+3(16u)$

$\\ \large\sf\longmapsto 53u+48u$

$\\ \large\sf\longmapsto 101u$

$\\ \large\sf\longmapsto 101g/mol$

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$\boxed{\sf No\:of\:moles=\dfrac{Given\:mass}{Molar\:mass}}$

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$\boxed{\sf Molarity=\dfrac{Moles\:of\:solute}{Vol\:of\:Solution\:in\:L}}$

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