Question

Составьте уравнение химической реакции и поставьте коэффициенты. Сколько литров сернистого газа образуется при сжигании 12,8 г кристаллической серы?

Answer 1

S+O2-->SO2 n(S)=12.8/32=0.4 моль
n(SO2)=0.4 моль
V(SO2)=0.4*22.4=8.96л

Answer 2

Chemical Reaction Equation and Coefficients

The chemical equation for the combustion of sulfur can be represented as:

S(s) + O2(g) → SO2(g)

To balance this equation, we need to ensure that the number of atoms of each element is the same on both sides of the equation. The balanced equation with coefficients is:

S(s) + 2O2(g) → 2SO2(g)

Calculation of Sulfur Dioxide Gas Produced

The molar mass of sulfur (S) is approximately 32.07 g/mol, and the molar mass of sulfur dioxide (SO2) is approximately 64.07 g/mol. Using these molar masses, we can calculate the amount of sulfur dioxide produced when 12.8 g of sulfur is burned.

First, we need to convert the mass of sulfur to moles using the molar mass of sulfur: 12.8 g / 32.07 g/mol = 0.399 moles of S

According to the balanced chemical equation, 1 mole of sulfur produces 2 moles of sulfur dioxide. Therefore, the moles of sulfur dioxide produced is: 0.399 moles of S × 2 moles of SO2 / 1 mole of S = 0.798 moles of SO2

Finally, we can convert moles of sulfur dioxide to liters at standard temperature and pressure (STP), where 1 mole of any gas occupies 22.4 liters at STP: 0.798 moles of SO2 × 22.4 L/mol = 17.87 liters of SO2

So, 17.87 liters of sulfur dioxide gas are produced when 12.8 g of crystalline sulfur is burned.