What is the maximum mass of S8 that can be produced by combining 93.0 grams of each reactant?
8SO2 + 16H2S ——> 3S8+16H2O
To find the maximum mass of S8 that can be produced, we need to determine the limiting reactant in the chemical reaction and calculate the amount of S8 that can be formed based on the stoichiometry of the reaction.
1. Start by calculating the molar masses of each reactant:
- Molar mass of SO2 (sulfur dioxide): 32.07 g/mol + 2*(16.00 g/mol) = 64.07 g/mol
- Molar mass of H2S (hydrogen sulfide): 2*(1.01 g/mol) + 32.07 g/mol = 34.08 g/mol
2. Convert the given mass of each reactant to moles:
- Moles of SO2 = 93.0 g / 64.07 g/mol = 1.45 mol
- Moles of H2S = 93.0 g / 34.08 g/mol = 2.73 mol
3. Using the coefficients from the balanced equation, determine the stoichiometric ratio of S8 to each reactant. In this case, the ratio of S8 to SO2 is 3:8, and the ratio of S8 to H2S is 3:16.
4. Calculate the maximum mass of S8 that can be formed from each reactant:
- Maximum mass of S8 from SO2 = (1.45 mol SO2) * (3 mol S8 / 8 mol SO2) * (256.52 g/mol) = 172.47 g
- Maximum mass of S8 from H2S = (2.73 mol H2S) * (3 mol S8 / 16 mol H2S) * (256.52 g/mol) = 160.31 g
5. Compare the two amounts and determine the limiting reactant. The limiting reactant is the one that produces a smaller amount of S8, as it will be completely consumed during the reaction. In this case, the limiting reactant is H2S, as it produces a smaller amount of S8.
Therefore, the maximum mass of S8 that can be produced by combining 93.0 grams of each reactant is 160.31 grams.
Looking at the mass relationships
8SO2>>8*64=512
16H2S=544 check those.
so now is to figure out how much of each divides the 93grams
SO2: 512/(1056)=.49*93=45.6 grams
H2S: the rest or 54.2 grams
Now how much S is that: 45.6*32/64=22.8 grams
plus 54.2*32/34=51.0 grams
add those (total S)=73.8 grams
check my calxulations