Ask your own question, for FREE!
Chemistry 18 Online
OpenStudy (unbelievabledreams):

I need help with Chemistry.

OpenStudy (unbelievabledreams):

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

OpenStudy (unbelievabledreams):

@Photon336

OpenStudy (photon336):

Will help out in a bit!

OpenStudy (photon336):

First step: Balance your equation: 2Al(OH)3 + 3H2SO4 ------> 6H2O + Al2(SO4)3

OpenStudy (photon336):

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

OpenStudy (unbelievabledreams):

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.

OpenStudy (photon336):

Sure. Give me about 45 min!

OpenStudy (unbelievabledreams):

Thanks a bunch.

OpenStudy (photon336):

Take a look at what I wrote, it's a little messy but ask questions.

OpenStudy (unbelievabledreams):

@Photon336 I am back. So the question 1 is 0.255 moles?

OpenStudy (photon336):

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.

OpenStudy (photon336):

In this case H2SO4 is your limiting reagent and you have 0.255 moles of that

OpenStudy (photon336):

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.

OpenStudy (unbelievabledreams):

I see.

OpenStudy (photon336):

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.

OpenStudy (photon336):

this will help you to solve part 2 of the question

OpenStudy (unbelievabledreams):

I got it.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!