25 moles of sulfur dioxidecontain 800,825 g oxygen.
Sulfur dioxide (SO2) consists of one sulfur atom and two oxygen atoms. The percentage of sulfur in sulfur dioxide is calculated as the mass of sulfur divided by the total mass of the compound, which is 32.07 grams per mole for sulfur and 64.07 grams per mole for sulfur dioxide. This means that sulfur represents 50% of the total mass of sulfur dioxide.
If 12 grams of carbon were used to form the 22 grams of carbon dioxide, this implies that 12 grams of oxygen were consumed in the reaction. Since 20 grams of oxygen were initially available, only 8 grams of oxygen are left unused.
The density of sulfur dioxide at standard temperature and pressure (STP) is approximately 2.927 grams per liter.
The gram-atomic mass of sulphur is 32 and that of oxygen is 16, to two significant digits. Therefore, the mass of oxygen with the same number of atoms as 64 grams of sulphur can be found from the proportion m/64 = 16/32, or m = 32 grams.
Sulfur is approximately 16 times heavier than oxygen. This is because sulfur has an atomic mass of around 32 grams per mole, while oxygen has an atomic mass of around 16 grams per mole.
800 g oxygen are needed.
Sulfur dioxide (SO2) consists of one sulfur atom and two oxygen atoms. The percentage of sulfur in sulfur dioxide is calculated as the mass of sulfur divided by the total mass of the compound, which is 32.07 grams per mole for sulfur and 64.07 grams per mole for sulfur dioxide. This means that sulfur represents 50% of the total mass of sulfur dioxide.
Sulfur has relative atomic mass of 32 and oxygen have that of 16. The molar mass of sulfur dioxide is 64 grams per mole. Therefore there is approximately 0.58 moles (37.14/64) of sulfur dioxide in given weight.
A sulfur dioxide has one sulfur atom and two oxygen atoms. Therefore, considering a mole of sulfur dioxide (64g); there is 32g of sulfur and 32g of oxygen. Hence the mass percent of oxygen is 50%.
If 12 grams of carbon were used to form the 22 grams of carbon dioxide, this implies that 12 grams of oxygen were consumed in the reaction. Since 20 grams of oxygen were initially available, only 8 grams of oxygen are left unused.
The formula for sulfate is SO4, which indicates that there is one sulfur atom for every four oxygen atoms (1:4) ratio. If this isn't what you're looking for, then perhaps you're looking for specific percentages. In that case, you need the atomic masses of the elements involved to find the total molecular mass.Sulfur = 32.1 gramsOxygen = 16.0 grams × 4 atoms = 64.0 grams----------------------------------------------------------Sulfate = 96.1 gramsThen you take the mass of each and divide it by the total mass to get a percentage.32.1 grams ÷ 96.1 grams = .334 = 33.4% sulfur in sulfate64.0 grams ÷ 96.1 grams = .666 = 66.6% oxygen in sulfate
11 grams because all is reacted and there is no reactant left over, although if there were only 3 grams of carbon there would have to be 6 grams of oxygen for this to be viable as carbon dioxide is CO2 so the question asked was itself wrong.
The density of sulfur dioxide at standard temperature and pressure (STP) is approximately 2.927 grams per liter.
Just about 6.022 X 1023 atoms of sulfur. Sulfur is 32.07 grams per mole.
Sulfur is a non metal element. Atomic mass of it is 32.
The gram-atomic mass of sulphur is 32 and that of oxygen is 16, to two significant digits. Therefore, the mass of oxygen with the same number of atoms as 64 grams of sulphur can be found from the proportion m/64 = 16/32, or m = 32 grams.
1 mol Sulfur is 32 g Sulfur So 2.5 mol Sulfur is 80 g Sulfur