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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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2. When do chemical reactions happen?

Alenkinab [10]

Either the 1or3answer

7 0

3 years ago

(2 KClO3 (s) → 2 KCl (s) + 3 O2 (g) ) If 165 mL of oxygen is produced at 30.0 °C and 90.0 kPa, what mass of KClO3 was decomposed

soldier1979 [14.2K]

Taking into account the reaction stoichiometry and ideal gas law, 0.48144 grams of KClO₃ was decomposed.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 KClO₃ → 2 KCl + 3 O₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

KClO₃: 2 moles
KCl: 2 moles
O₂: 3 moles

The molar mass of the compounds is:

KClO₃: 122.45 g/mole
KCl: 74.45 g/mole
O₂: 32 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

KClO₃: 2 moles ×122.45 g/mole= 244.8 grams
KCl: 2 moles ×74.45 g/mole= 148.9 grams
O₂: 3 moles ×32 g/mole= 96 grams

<h3>Ideal gas law</h3>

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P×V = n×R×T

where:

P is the gas pressure.
V is the volume that the gas occupies.
T is the temperature of the gas.
R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.
n is the number of moles of the gas.

<h3>Number of O₂ produced.</h3>

165 mL of oxygen is produced at 30.0 °C and 90.0 kPa. This is, you know:

P= 90 kPa= 0.888231 atm (being 101.325 kPa= 1 atm)
V= 165 mL= 0.165 L (being 1000 mL= 1 L)
n= ?
R= 0.082 $\frac{atmL}{molK}$
T= 30 C= 303 K (being 0 C= 273 K)

Replacing in the ideal gas law:

0.888231 atm× 0.165 L = n× 0.082 $\frac{atmL}{molK}$ × 303 K

Solving:

n= (0.888231 atm× 0.165 L)÷ (0.082 $\frac{atmL}{molK}$ × 303 K)

<u><em>n= 0.0059 moles</em></u>

Finally, 0.0059 moles of oxygen is produced at 30 °C and 90 kPa.

<h3>Mass of KClO₃ required</h3>

The following rule of three can be applied: If by stoichiometry of the reaction 3 moles of O₂ are produced by 244.8 grams of KClO₃, 0.0059 moles of O₂ are produced by how much mass of KClO₃?

$mass of KClO_{3}= \frac{0.0059 moles of O_{2}x 244.8 grams of KClO_{3}}{3 moles of O_{2}}$