2SO2(g) + O2(g) >2SO3(g). If 96g SO2 is added to 2 moles of oxygen at STP, calculate the mass of SO3 that is formed
This is a limiting reagent problem. You know that because amounts are given for BOTH reactants.
2SO2 + O2 ==> 2SO3
mols sO2 = grams/molar mass
mols O2 = 2
Using the coefficients in the balanced equation convert mols SO2 to mols SO3.
Do the same for mols O2 to mols SO3.
It is likely that these two values will not agree which means one of them is wrong. The correct value in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
Now convert mols SO3 to grams. g = mols x molar mass.
The molar mass of SO3 is 80g.mol to the power -1 and we have 2 moles...so the mass is 80g.mol to the power -1 x 2 mol..the answer will be 160 g
Calculate the mass and volume of oxygen released when 29.4g pottasium chlorate decomposes to form pottasium chloride and oxygen
To calculate the mass of SO3 formed, we need to use stoichiometry. Stoichiometry is a relationship between the amounts of reactants and products in a balanced chemical equation.
First, let's determine the molar mass of SO2 (sulfur dioxide), O2 (oxygen), and SO3 (sulfur trioxide):
- Molar mass of SO2 = 32.06 g/mol
- Molar mass of O2 = 31.998 g/mol (approximately 32 g/mol)
- Molar mass of SO3 = 80.06 g/mol
Now, let's use the balanced chemical equation to find the molar ratio between SO2 and SO3:
2SO2(g) + O2(g) → 2SO3(g)
From the equation, we can see that 2 moles of SO2 react to form 2 moles of SO3. This means the molar ratio between SO2 and SO3 is 1:1.
Given that we have 96 g of SO2, we need to convert it to moles using the molar mass of SO2:
Moles of SO2 = mass / molar mass
Moles of SO2 = 96 g / 32.06 g/mol = 2.995 mol (approximately 3 mol)
Since the molar ratio between SO2 and SO3 is 1:1, the moles of SO3 formed will be the same as the moles of SO2:
Moles of SO3 = 2.995 mol
Now, we can calculate the mass of SO3 using its molar mass:
Mass of SO3 = moles of SO3 * molar mass of SO3
Mass of SO3 = 2.995 mol * 80.06 g/mol = 239.8 g (approximately 240 g)
Therefore, approximately 240 grams of SO3 will be formed when 96 grams of SO2 reacts with 2 moles of oxygen at STP.