What is the mass of gas mixture containing 95.5 O2 required for it to react with an excess of Cs2 to produce 6.83g of So2?
The mass of the gas mixture containing 95.5 O2 necessary for it to react with an excess of Cs2 to produce 6.83g of So2 is approximately 5.1232 grams.
Calculating the Mass of Gas Mixture Containing 95.5 O2
To calculate the mass of the gas mixture containing 95.5 O2 necessary for the reaction, we need to follow these steps:
1. Convert the given mass of So2 to moles. The molar mass of So2 is 64.06 g/mol.
2. Use the mole ratio from the balanced equation to determine the moles of O2 required.
3. Convert the moles of O2 to grams using the molar mass of O2.
Step 1: Convert the mass of So2 to moles:
Mass of So2 = 6.83 g
Molar mass of So2 = 64.06 g/mol
Number of moles of So2 = Mass of So2 / Molar mass of So2
Number of moles of So2 = 6.83 g / 64.06 g/mol = 0.1067 mol
Step 2: Use the mole ratio to determine the moles of O2 required:
From the balanced equation Cs2 + 3O2 -> Co2 + 2So, we know that 2 moles of So2 are produced from 3 moles of O2.
Number of moles of O2 = (Number of moles of So2) x (3 moles of O2 / 2 moles of So2)
Number of moles of O2 = 0.1067 mol x (3 mol O2 / 2 mol So2) = 0.1601 mol
Step 3: Convert the moles of O2 to grams:
Molar mass of O2 = 32 g/mol
Mass of O2 = (Number of moles of O2) x (Molar mass of O2)
Mass of O2 = 0.1601 mol x 32 g/mol = 5.1232 g
This calculation shows that approximately 5.1232 grams of the gas mixture containing 95.5 O2 are necessary for the reaction to produce 6.83g of So2.