The combustion of carbon disulfide in the presence of excess oxygen yields carbon dioxide and sulfur dioxide:
CS2(g)+3O2(g)--->CO2(g)+2SO2(g)
The combustion of 17.2 g of CS2 in the presence of excess oxygen yields __________ g of SO2.
See your post above.
25.3
23
To find the mass of SO2 produced, we need to use stoichiometry - the relationship between the reactants and products in a chemical reaction.
The balanced equation for the reaction tells us that 1 mole of CS2 reacts with 2 moles of SO2. First, we need to find the number of moles of CS2 present in 17.2 g.
To do this, we can use the molar mass of CS2, which is 76.14 g/mol. We divide the mass of CS2 by the molar mass to get moles:
Moles = Mass / Molar mass
Moles = 17.2 g / 76.14 g/mol ≈ 0.226 moles
According to the balanced equation, 1 mole of CS2 reacts to produce 2 moles of SO2. Therefore, the number of moles of SO2 produced will be twice the number of moles of CS2:
Moles of SO2 = 2 * Moles of CS2
Moles of SO2 = 2 * 0.226 moles = 0.452 moles
Now we can determine the mass of SO2 produced by multiplying the number of moles by the molar mass of SO2. The molar mass of SO2 is 64.06 g/mol:
Mass = Moles * Molar mass
Mass = 0.452 moles * 64.06 g/mol ≈ 28.9 g
Therefore, the combustion of 17.2 g of CS2 in the presence of excess oxygen yields approximately 28.9 g of SO2.