a chemist wanted to make 100 grams of SO2. She had only 12 grams of sulfur and 4 grams of oxygen. Unfortunately, the amounts of reactants were not enough to make the desired product. How much S and O2 would she need in order to obtain the desired grams of product with no leftovers?
S + 02 = SO2 + NO LEFT OVER
See your post above.
To determine the amounts of sulfur (S) and oxygen gas (O2) needed to produce 100 grams of sulfur dioxide (SO2) with no leftovers, we need to use stoichiometry.
1. Write the balanced chemical equation for the reaction:
S + O2 → SO2
2. Determine the molar masses of the compounds involved:
- Molar mass of S: 32 g/mol
- Molar mass of O2: 32 g/mol (16 g/mol per oxygen atom)
- Molar mass of SO2: 64 g/mol (32 g/mol per sulfur atom and 16 g/mol per oxygen atom)
3. Calculate the number of moles of the given reactants:
- Moles of S: 12 g / 32 g/mol = 0.375 mol
- Moles of O2: 4 g / 32 g/mol = 0.125 mol
4. Determine the limiting reactant:
The limiting reactant is the one that is completely consumed in the reaction and determines the maximum amount of product that can be formed. To find the limiting reactant, compare the mole ratios between the reactants and the product.
According to the balanced equation, the mole ratio between S and SO2 is 1:1, and the mole ratio between O2 and SO2 is 1:1. Since the mole ratio of S to SO2 is the same as that for O2 to SO2, we can compare the number of moles for both reactants directly.
0.375 mol S → 0.375 mol SO2
0.125 mol O2 → 0.125 mol SO2
Since the number of moles of O2 (0.125 mol) is less than the number of moles of S (0.375 mol), O2 is the limiting reactant.
5. Determine the amount of the excess reactant:
Subtract the moles of the limiting reactant (O2) from the moles of the excess reactant (S) to find the moles that are left over.
Moles left over = Moles of S - Moles of O2 = 0.375 mol - 0.125 mol = 0.250 mol
6. Calculate the mass of SO2 formed from the limiting reactant:
To determine the mass of SO2 formed from the limiting reactant, multiply the moles of the limiting reactant by the molar mass of SO2.
Mass of SO2 = Moles of O2 × Molar mass of SO2 = 0.125 mol × 64 g/mol = 8 grams
7. Calculate the total mass of SO2 required:
Since the desired mass of SO2 is 100 grams, we subtract the mass of SO2 formed from the limiting reactant from the desired mass.
Total mass of SO2 required = Desired mass of SO2 - Mass of SO2 formed from limiting reactant = 100 g - 8 g = 92 grams
8. Determine the amounts of S and O2 needed to produce the desired amount of SO2:
The mole ratio between S and SO2 is 1:1, so the moles of S needed to produce the desired mass of SO2 are the same as the moles of SO2 required. Multiply the moles of S by the molar mass of S to find the mass of S needed.
Mass of S needed = Moles of S × Molar mass of S = 0.125 mol × 32 g/mol = 4 grams
Since the mole ratio between O2 and SO2 is 1:1, the moles of O2 needed to produce the desired mass of SO2 are also the same as the moles of SO2 required. Multiply the moles of O2 by the molar mass of O2 to find the mass of O2 needed.
Mass of O2 needed = Moles of O2 × Molar mass of O2 = 0.125 mol × 32 g/mol = 4 grams
So, to obtain the desired 100 grams of SO2 with no leftovers, the chemist would need 4 grams of sulfur (S) and 4 grams of oxygen gas (O2).