How many grams of aluminum sulfate, Al2( SO4)3 , will be formed from 0.98 g of sulfuric acid and enough aluminum according to the following reaction:
H2SO4 + Al -> Al2(SO4)3 + H2
Follow the steps in this worked example.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To find the number of grams of aluminum sulfate formed, we need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed and determines the amount of product formed.
1. Start by calculating the molar mass of sulfuric acid (H2SO4):
H2: 2(1.01 g/mol) = 2.02 g/mol
S: 1(32.07 g/mol) = 32.07 g/mol
O4: 4(16.00 g/mol) = 64.00 g/mol
Total: 2.02 + 32.07 + 64.00 = 98.09 g/mol
2. Next, convert the given mass of sulfuric acid to moles:
Moles of H2SO4 = mass / molar mass
Moles of H2SO4 = 0.98 g / 98.09 g/mol = 0.01 mol
3. According to the balanced chemical equation, the stoichiometric ratio between sulfuric acid and aluminum sulfate is 1:1. It means that for every 1 mole of sulfuric acid, we will produce 1 mole of aluminum sulfate.
4. To find the moles of aluminum sulfate formed, the moles of sulfuric acid can be used since they are equal:
Moles of Al2(SO4)3 = 0.01 mol
5. Finally, calculate the mass of aluminum sulfate using its molar mass:
Molar mass of Al2(SO4)3:
Al2: 2(26.98 g/mol) = 53.96 g/mol
S3: 3(32.07 g/mol) = 96.21 g/mol
O12: 12(16.00 g/mol) = 192.00 g/mol
Total: 53.96 + 96.21 + 192.00 = 342.17 g/mol
Mass of Al2(SO4)3 = moles × molar mass
Mass of Al2(SO4)3 = 0.01 mol × 342.17 g/mol = 3.42 g
Therefore, 3.42 grams of aluminum sulfate (Al2(SO4)3) will be formed from 0.98 grams of sulfuric acid and enough aluminum.