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1. When 25.6 g of Al(OH)3 and 25.6 g of H2SO4 are reacted, how many moles of Al2(SO4)3 can be produced? 2. When 55.8 g of H2SO4 and 35.8 g of Al(OH)3 are reacted, how many grams of Al2(SO4)3 are produced? 3. In a reaction vessel, 3.4 mol of Al(OH)3 and 7.3 mol of H2SO4 react. Determine which substances, and their molar quantities, will be present in the container when the reaction goes to completion. Moles of Al2(SO4)3 in container: Moles of H2O in container: Moles of excess reactant in container: 2 Al(OH)3 + 3 H2SO4 ---> Al2(SO4)3 + 6 H2O <--- Balanced equation
Will help out in a bit!
First step: Balance your equation: 2Al(OH)3 + 3H2SO4 ------> 6H2O + Al2(SO4)3
In this problem you may have something called a limiting reagent. The limiting reagent is the compound that you have fewer moles for and runs out first. So you use the number of moles of the limiting reagent to find how many moles of Product you have. But first try to find how many moles of al(OH)3 and h2SO4 you have
Can you tell me all of these steps? I will be back in an hour because of my grandfather. Please help me. Thank you so much.
Sure. Give me about 45 min!
Thanks a bunch.
@Photon336 I am back. So the question 1 is 0.255 moles?
oh that sheet I wrote 2) on the side. once you figure out how many moles of reactants you have, you need to multiply each but the molar ratio to figure out which substance is limiting.
In this case H2SO4 is your limiting reagent and you have 0.255 moles of that
You need to use your limiting reagent to figure out how many moles of products you have, that's why it's so important, and you must always identity it first.
on the paper that's the purpose of step 2. you can clearly see that Aluminum is in excess while H2SO4 is not and all i did was multiply the number of moles of each by the molar ratio as it appears in the balanced equation.
this will help you to solve part 2 of the question
I got it.