Question

Определите объем образующегося газа, если с серой массой 120 г реагировал кислород.

Answer 1

Дано:

m(S)=120г

Найти:

V(SO2↑)-?

Решение:

120г---------------х г

S + O2 = SO2

1моль------------1моль

32г/моль-----64г/моль

М=32г-----------М=64г

найдём х:

$\frac{120}{32} = \frac{x}{64}$

$x = \frac{120 \times 64}{32}$

$x = 240$

Найдем:V(SO2)

V=m/M×Vm=240/64×22,4=84,7л

Ответ: V(SO2↑)=84,7 л

Answer 2

Calculating the Volume of Gas Produced

To calculate the volume of gas produced when 120g of sulfur reacts with oxygen, we can use the balanced chemical equation for the reaction. The balanced chemical equation for the reaction of sulfur with oxygen to produce sulfur dioxide is:

S(s) + O2(g) → SO2(g)

From the balanced chemical equation, we can see that 1 mole of sulfur reacts with 1 mole of oxygen to produce 1 mole of sulfur dioxide. We can use the molar mass of sulfur to convert the given mass of sulfur to moles and then use the mole ratio from the balanced equation to find the moles of oxygen consumed and the moles of sulfur dioxide produced. Finally, we can use the ideal gas law to calculate the volume of sulfur dioxide produced.

Calculation Steps

1. Calculate the moles of sulfur using its molar mass. 2. Use the mole ratio from the balanced equation to find the moles of oxygen consumed and the moles of sulfur dioxide produced. 3. Apply the ideal gas law to calculate the volume of sulfur dioxide produced.

Molar Mass of Sulfur

The molar mass of sulfur (S) is approximately 32.07 g/mol.

Calculation

1. Moles of sulfur: - Molar mass of sulfur = 32.07 g/mol - Moles of sulfur = Mass of sulfur / Molar mass of sulfur - Moles of sulfur = 120g / 32.07 g/mol - Moles of sulfur ≈ 3.74 mol

2. Moles of oxygen and sulfur dioxide: - From the balanced equation, 1 mole of sulfur reacts with 1 mole of oxygen to produce 1 mole of sulfur dioxide. - Therefore, moles of oxygen = 3.74 mol - Moles of sulfur dioxide = 3.74 mol

3. Volume of sulfur dioxide: - Using the ideal gas law: PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is the temperature. - Assuming standard temperature and pressure (STP), where T = 273 K and P = 1 atm. - The molar volume of a gas at STP is approximately 22.4 L/mol. - Therefore, volume of sulfur dioxide = Moles of sulfur dioxide * Molar volume at STP - Volume of sulfur dioxide ≈ 3.74 mol * 22.4 L/mol - Volume of sulfur dioxide ≈ 83.776 L

Conclusion

The volume of sulfur dioxide produced when 120g of sulfur reacts with oxygen is approximately 83.776 liters.