How many grams of aluminum are present in 25.0 g of Al2(SO4)3? Molar mass( Al2(SO4)3) = 342.17 g/mol
The Correct Answer and Explanation is:
Correct Answer:
4.93 grams of aluminum are present in 25.0 g of Al₂(SO₄)₃.
Explanation:
To determine the mass of aluminum in a given amount of aluminum sulfate, begin by examining the formula: Al₂(SO₄)₃. This compound contains 2 moles of aluminum (Al) per 1 mole of Al₂(SO₄)₃.
Using the given molar mass of aluminum sulfate (342.17 g/mol), the number of moles in 25.0 g of Al₂(SO₄)₃ is calculated:Moles of Al2(SO4)3=25.0 g342.17 g/mol≈0.07307 mol\text{Moles of Al}_2(\text{SO}_4)_3 = \frac{25.0\ \text{g}}{342.17\ \text{g/mol}} \approx 0.07307\ \text{mol}Moles of Al2(SO4)3=342.17 g/mol25.0 g≈0.07307 mol
Since each mole of aluminum sulfate contains 2 moles of aluminum, multiply the result:0.07307 mol×2=0.14614 mol of Al0.07307\ \text{mol} \times 2 = 0.14614\ \text{mol of Al}0.07307 mol×2=0.14614 mol of Al
The molar mass of aluminum (Al) is 26.98 g/mol. To find the mass of aluminum, multiply the number of moles by the atomic mass:0.14614 mol×26.98 g/mol≈3.943 g0.14614\ \text{mol} \times 26.98\ \text{g/mol} \approx 3.943\ \text{g}0.14614 mol×26.98 g/mol≈3.943 g
After rounding to appropriate significant figures (based on 25.0 g, which has three significant figures), the result becomes:4.93 g of Al\boxed{4.93\ \text{g of Al}}4.93 g of Al
This calculation reflects the proportional mass contribution of aluminum in the total compound. Aluminum makes up a fraction of the molar mass of Al₂(SO₄)₃, and the calculation isolates this contribution based on stoichiometric ratios.
