23. aluminum reacts with sulfuric acid to produce aluminum sulfate and hydrogen gas. how many grams of aluminum sulfate would be formed if 250g H2SO4 reacted with aluminum?
To calculate the grams of aluminum sulfate formed, we need to use the balanced chemical equation for the reaction between aluminum and sulfuric acid.
The balanced equation is:
2 Al + 3 H2SO4 -> Al2(SO4)3 + 3 H2
According to the equation, for every 3 moles of H2SO4, we get 1 mole of Al2(SO4)3.
First, let's convert the given mass of H2SO4 to moles:
Molar mass of H2SO4 = 2 * (1.01 g/mol) + 32.07 g/mol + 4 * (16.00 g/mol) = 98.09 g/mol
Moles of H2SO4 = Mass / Molar mass = 250 g / 98.09 g/mol
Next, we need to determine the moles of Al2(SO4)3 formed based on the mole ratio from the balanced equation:
Moles of Al2(SO4)3 = Moles of H2SO4 / 3
Finally, we can calculate the mass of Al2(SO4)3 formed:
Mass of Al2(SO4)3 = Moles of Al2(SO4)3 * Molar mass of Al2(SO4)3
The molar mass of Al2(SO4)3 can be calculated as follows:
Molar mass of Al2(SO4)3 = 2 * (26.98 g/mol) + 3 * (32.07 g/mol) + 4 * (16.00 g/mol) = 342.15 g/mol
Substituting the values into the equation:
Moles of Al2(SO4)3 = (250 g / 98.09 g/mol) / 3
Mass of Al2(SO4)3 = (250 g / 98.09 g/mol) / 3 * 342.15 g/mol
Calculating this, we find:
Mass of Al2(SO4)3 = 277.62 grams
Therefore, approximately 277.62 grams of aluminum sulfate would be formed.
To find the number of grams of aluminum sulfate formed, we need to understand the stoichiometry of the reaction. The balanced chemical equation for the reaction is:
2 Al + 3 H2SO4 → Al2(SO4)3 + 3 H2
From the equation, we can see that 2 moles of aluminum (Al) react with 3 moles of sulfuric acid (H2SO4) to produce 1 mole of aluminum sulfate (Al2(SO4)3) and 3 moles of hydrogen gas (H2).
Now, we need to convert the given mass of sulfuric acid (H2SO4) to moles using its molar mass. The molar mass of H2SO4 is:
H2SO4:
2(1.01 g/mol H) + 32.07 g/mol S + 4(16.00 g/mol O) = 98.09 g/mol
So, 250 grams of H2SO4 is equal to:
250 g H2SO4 / 98.09 g/mol H2SO4 = 2.55 moles H2SO4
From the balanced chemical equation, we can see that the mole ratio between H2SO4 and Al2(SO4)3 is 3:1. Therefore, for every 3 moles of H2SO4, 1 mole of Al2(SO4)3 is produced.
Hence, for 2.55 moles of H2SO4, the number of moles of Al2(SO4)3 formed would be:
2.55 moles H2SO4 * (1 mole Al2(SO4)3 / 3 moles H2SO4) = 0.85 moles Al2(SO4)3
Finally, to find the mass of aluminum sulfate formed, we multiply the number of moles by its molar mass:
0.85 moles Al2(SO4)3 * (342.15 g/mol Al2(SO4)3) ≈ 291.33 grams
Therefore, approximately 291.33 grams of aluminum sulfate would be formed if 250 grams of H2SO4 reacted with aluminum.
1. Write the equation.
2. Convert 250 g H2SO4 to mols. mols = g/molar mass.
3. Convert mols H2SO4 to mols Al2(SO4)3 using the coefficients in the balanced equation.
4. Convert mols Al2(SO4)3 to grams. g = mols x molar mass.
4.