Final answer:
To determine the mass of Al2(SO4)3 produced, the moles of H2SO4 are first calculated using the volume and molarity, then converted to moles of Al2(SO4)3 using the balanced chemical equation, and finally multiplied by the formula mass to find the mass in grams.
Step-by-step explanation:
The question asks about the mass of Al2(SO4)3 that can be produced from a given volume and concentration of H2SO4 solution. To solve this, we first need to calculate the number of moles of H2SO4 by multiplying the volume (in liters) by the concentration of the solution.
Since the concentration of H2SO4 is 0.200 M, we find the number of moles as follows:
0.355 L * 0.200 mol/L = 0.071 mol H2SO4
Using the balanced chemical equation 2 Al(s) + 3H2SO4 (aq) → Al2(SO4)3 (aq) + 3H2, we see that it takes 3 moles of H2SO4 to produce 1 mole of Al2(SO4)3. Thus, the number of moles of Al2(SO4)3 produced will be 0.071 mol H2SO4 ÷ 3 = 0.0237 mol Al2(SO4)3.
The formula mass for Al2(SO4)3 is 342.14 amu, so the mass of Al2(SO4)3 will be:
0.0237 mol * 342.14 g/mol ≈ 8.11 g
Therefore, the mass of Al2(SO4)3 produced from 355 mL of 0.200 M H2SO4 is approximately 8.11 grams.