Answer: The atomic number
The balanced chemical reaction is expressed as:
<span>8SO2 + 16H2S = 3S8 + 16H2O
We are given the initial amount of the reactants. From there, we determine the limiting reactant. We do as follows:
87.0 g SO2 ( 1 mol / 64.07 g ) = 1.36 mol SO2 ( 16 mol H2S / 8 mol SO2 ) = 2.72 mol H2S
87.0 g H2S ( 1 mol / 34.08 g ) = 2.55 mol H2S ( 8 mol SO2 / 16 mol H2S ) = 1.28 mol SO2
Therefore, the limiting reactant would be H2S. We calculate the maximum amount of S8 that can be produced from the amount of H2S.
2.55 mol H2S ( 3 mol S8 / 16 mol H2S ) ( 256.48 g / 1 mol ) = 122.63 g S8</span>
Answer:
d. 12.3 grams of Al2O3
Explanation:
The balanced chemical equation of this chemical reaction is as follows:
4Al + 3O2 --> 2Al2O3
Based on the balanced equation, 4 moles of aluminum (limiting reagent) reacts to form 2 moles of aluminum oxide (Al2O3).
First, we need to convert the mass of aluminum to moles using the formula;
mole = mass/molar mass
Molar mass of Al = 27g/mol
mole = 6.50/27
= 0.241mol of Al.
Hence, if 4 moles of aluminum (limiting reagent) reacts to form 2 moles of aluminum oxide (Al2O3).
Then, 0.241mol of Al will produce (0.241 × 2/4) = 0.241/2 = 0.121mol of Al2O3.
Convert this mole value to molar mass using mole = mass/molar mass
Molar mass of Al2O3 = 27(2) + 16(3)
= 54 + 48
= 102g/mol
mass = molar mass × mole
mass = 102 × 0.121
mass of Al2O3 = 12.34grams.