Answer: We must first compute the quantity of H2O that can result from the reaction of AL(CH)3 and H2SO4 in order to determine the mass of H2O produced.
Step-by-step explanation:
Let's begin by formulating the reaction's balanced chemical equation:
Al2(SO4)3 + 6 H2O + 2 CH4 = 2 AL(CH)3 + 3 H2SO4
According to the equation, 2 moles of AL(CH)3 and 3 moles of H2SO4 combine to form 6 moles of H2O.
1.75 g of AL(CH)3 is equal to: 1.75 g / 78.0 g/mol = 0.0224 mol of AL(CH)3 as 1 mole of AL(CH)3 has a molar mass of roughly 78.0 g/mol.
2.00 g of H2SO4 is equal to: 2.00 g / 98.1 g/mol = 0.0204 mol of H2SO4 as 1 mole of H2SO4 has a molar mass of roughly 98.1 g/mol.
In order to calculate the amount of H2O produced, we must now identify the limiting reagent.
We contrast the moles of AL(CH)3 and H2SO4 in use. AL(CH)3 and H2SO4 have a stoichiometric ratio of 2 to 3. As a result, 3 moles of H2SO4 are required for every 2 moles of AL(CH).
The equation states that 2 moles of AL(CH)3 will yield 6 moles of water. Therefore, we can calculate the moles of H2O created for 0.0224 mol of AL(CH)3 as follows: 0.0224 mol AL(CH)3 (6 mol H2O / 2 mol AL(CH)3) = 0.0672 mol H2O
Finally, we may use the molar mass of H2O to convert its moles to grams. H2O has a molar mass of about 18.0 g/mol:
18.0 g/mol x 0.0672 mol H2O = 1.21 g H2O
As a result, the mass of H2O generated is roughly 1.21 grams if H2SO4 is the limiting reagent.