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mariarad [96]
3 years ago
15

3. Theoretically how many grams of magnesium is required to produce to 5.0 g of

Chemistry
1 answer:
FinnZ [79.3K]3 years ago
8 0

Answer:

3grams

Explanation:

The reaction for the production of Magnesium dioxide will be

Mg + O2  → MgO

we have 5g of MgO (molar mass 40g)

no of moles of MgO = 5/40 = 0.125

Using unitary method we have

1 mole of Mg require 1 mole of MgO

0.125 Mole of MgO = 0.125mole of Mg

n = given mass /molar mass

0.125 = mass / molar mass

mass = 0.125* 24 = 3grams

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Answer: 100C of heat is needed.

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Sodas are made fizzy by the carbon dioxide ( CO ) dissolved in the liquid. An amount of carbon 2 dioxide equal to about 8.0 L of
eduard

Answer:

A) The number of moles of Carbon dioxide in 2.0 L bottle of soda = 0.325 mole

B) The mass of Carbon dioxide in 2.0 L bottle of soda = 14.3 g

Explanation:

Complete Question

Soda are made fizzy by the carbon dioxide (CO₂) dissolved int he liquid. An amount of carbon dioxide equal to about 8.0 L of carbon dioxide gas at atmospheric pressure and 300.0 K can be dissolved can be dissolved in a 2-L bottle of soda. The molar mass of CO₂ IS 44 g/mol.

A) How many moles of carbon dioxode are in the 2-L? (1L = 0.001 m³)

B) What is the mass of the carbon dioxide in the 2-L bottle of soda?

Solution

Assuming that CO₂ is an ideal gas.

A) The amount of carbon dioxide in 2.0 L of soda bottle is equal to the amount of Carbon dioxide in about 8.0 L of carbon dioxide gas at atmospheric pressure and 300.0 K

From the ideal gas equation, we know that

PV = nRT

P = pressure of the gas = atmospheric pressure = 101,325 Pa

V = volume of the gas = 8.0 L = 0.008 m³

n = number of moles of the ideal gas = ?

R = molar gas constant = 8.314 J/mol.K

T = absolute temperature of the gas in Kelvin = 300 K

101325 × 0.008 = n × 8.314 × 300

n = (101325×0.008) ÷ (8.314 × 300)

n = 0.324993986 = 0.325 mole of Carbon dioxide.

B) The mass of Carbon dioxide in 2.0 L bottle of soda

Mass = (Number of moles) × (Molar mass)

= 0.325 × 44.0

= 14.3 g

Hope this Helps!!!

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