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2 years ago

4. How many grams of hydrogen is produced from 12.5 g of Mg reacting with hydrochloric acid in this

Chemistry

1 answer:

Volgvan2 years ago

8 0

From the reactions, 1.04 g of H2 and 7.995 g of aluminum phosphate is produced.

<h3>What is stoichiometry?</h3>

The term stoichiometry has to do with the amount of substances that participates in a reaction.

For reaction 1;

Mg + 2HCl → MgCl₂ + H₂

Number of moles of Mg reacted = 12.5 g/24g/mol = 0.52 moles

If 1 mole of Mg produced 1 mole of H2

0.52 moles produces 0.52 moles of H2

Mass of H2 = 0.52 moles * 2 g/mol = 1.04 g

For reaction 2;

2Li3PO4 + Al2(SO4)3 → 3Li2SO4 + 2AIPO4

Number of moles of lithium phosphate = 7.5 g/116 g/mol = 0.065 moles

2 moles of Li3PO4 produced 2 moles of AIPO4

0.065 moles of Li3PO4 produced 0.065 moles of AIPO4

Mass of AIPO4 = 0.065 moles * 123 g/mol = 7.995 g

Learn more about stoichiometry:brainly.com/question/9743981

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Ksivusya [100]

<h2>Hello!</h2>

The answer is:

We have that there were produced 0.120 moles of $CO_{2}$

$n=0.120mol$

We are asked to calculate the number of moles of the given gas, also, we are given the volume, the temperature and the pressure of the gas, we can calculate the approximate volume using The Ideal Gas Law.

The Ideal Gas Law is based on Boyle's Law, Gay-Lussac's Law, Charles's Law, and Avogadro's Law, and it's described by the following equation:

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Where,

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T is the absolute temperature of the gas (Kelvin).

R is the ideal gas constant (to work with pressure in mmHg), which is equal to:

$R=62.363\frac{mmHg.L}{mol.K}$

We must remember that the The Ideal Gas Law equation works with absolute temperatures (K), so, if we are given relative temperatures such as Celsius degrees or Fahrenheit degrees, we need to convert it to Kelvin before we proceed to work with the equation.

We can convert from Celsius degrees to Kelvin using the following formula:

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So, we are given the following information:

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Now, isolating the number of moles, and substituting the given information, we have:

$PV=nRT$

$n=\frac{PV}{RT}$

$n=\frac{760mmHg*2.965L}{62.363\frac{mmHg.L}{mol.K}*298.5K}$

$n=\frac{760mmHg*2.965L}{62.363\frac{mmHg.L}{mol.K}*298.5K}\\\\n=\frac{2242mmHg.L}{18615.355\frac{mmHg.L}{mol.}}\\\\n=0.120mole$

Hence, we have that there were produced 0.120 moles of $CO_{2}$

$n=0.120mol$

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