Question

Визначити об'єм кисню, який необхідний для спалювання 8 гр. магнію.

Answer 1

Дано:
m(Mg)=8г
Знайти:
V(O2)=?

Р-ня:

8г                     х л           
2Mg          +    O2 =               2MgO
2 моль*         1моль *      
*М(Mg)            *22,4 л/моль

М(Mg) = 24 г/моль

8 г/48 г = х л/22,4 л/моль
х =(8*22,4)/48 = 4 л (О2)

В-дь: V(O2) = 4 л

Answer 2

Calculating the Volume of Oxygen Required for Burning 8g of Magnesium

To calculate the volume of oxygen required for burning 8g of magnesium, we can use the chemical equation for the combustion of magnesium:

2Mg + O2 → 2MgO

From the balanced chemical equation, we can see that 2 moles of magnesium react with 1 mole of oxygen to produce 2 moles of magnesium oxide.

Using the molar mass of magnesium (Mg = 24.305 g/mol) and oxygen (O2 = 32 g/mol), we can calculate the amount of oxygen required.

The molar mass of magnesium oxide (MgO) is 40.3 g/mol.

Calculation

1. Calculate the moles of magnesium: - Given mass of magnesium = 8g - Moles of magnesium = Given mass / molar mass of magnesium - Moles of magnesium = 8g / 24.305 g/mol - Moles of magnesium = 0.329 moles

2. Calculate the moles of oxygen required: - From the balanced chemical equation, 2 moles of magnesium react with 1 mole of oxygen. - Moles of oxygen required = Moles of magnesium / 2 - Moles of oxygen required = 0.329 moles / 2 - Moles of oxygen required = 0.1645 moles

3. Calculate the volume of oxygen at standard conditions (STP): - At standard conditions (0°C and 1 atm), 1 mole of any gas occupies 22.4 liters. - Volume of oxygen required = Moles of oxygen required * 22.4 L/mol - Volume of oxygen required = 0.1645 moles * 22.4 L/mol - Volume of oxygen required = 3.68 liters

Therefore, the volume of oxygen required for burning 8g of magnesium at standard conditions is approximately 3.68 liters.

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