What is the minimum amount of 5.2M H2SO4 necessary to produce 22.8g of H2 (g) according to the following reaction? 2Al(s)+3H2SO4(aq)→Al2(SO4)3(aq)+3H2(g)
(Please be precise and clear)
1. You have the balanced equation.
2. How many mols H2 do you want? That's mols = grams/molar mass.
3. Using the coefficients in the balanced equation, convert mols H2 to mols H2SO4.
4. Then M = mols/L. You know M H2SO4 and mols H2SO4, solve for L H2SO4.
This four step method will work just about any strictly stoichiometry problem. With a slight modification it will also work limiting regent problems.
To solve this problem, you need to use stoichiometry and the molar ratios provided by the balanced chemical equation.
Step 1: Calculate the moles of H2 using the given mass.
Given Mass of H2: 22.8 g
Molar mass of H2: 2.016 g/mol
Moles of H2 = mass / molar mass
= 22.8 g / 2.016 g/mol
= 11.31 mol (rounded to two decimal places)
Step 2: Use the stoichiometry of the balanced equation to determine the moles of H2SO4.
From the balanced equation, you can see that for every 3 moles of H2SO4, 3 moles of H2 are produced. Therefore, the molar ratio of H2SO4 to H2 is 3:3 or 1:1.
To find the moles of H2SO4 required, it will be the same as the moles of H2:
Moles of H2SO4 = 11.31 mol
Step 3: Convert moles of H2SO4 to the mass of H2SO4.
Molar mass of H2SO4: 98.09 g/mol
Mass of H2SO4 = moles x molar mass
= 11.31 mol x 98.09 g/mol
= 1110.56 g (rounded to two decimal places)
So, the minimum amount of 5.2M H2SO4 necessary to produce 22.8g of H2 is 1110.56 grams.
Note: It's important to note that the concentration of the sulfuric acid solution (5.2M) is irrelevant to this calculation, as the Stoichiometry is only concerned with the number of moles and not the concentration.