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assume that 13.5g of AL react with HCl according to the following equation, at STP: Al(s) + HCl(aq) ---> AlCl3 + H2(g) Remember to balance the equation first.

a. how many moles of Al react?
b. how mang moles of H2 are produced?
c. how mang liters of H2 at STP are produced?

thank you :]

So many concepts here. What is it you don't understand?

everything my teacher didn't teach us

actually i just need c

In the future, if you will show what you can do and work through that, then ask your question, all of this preliminary stuff is not necessary.

Remember that a mole of any (ideal) gas will occupy 22.4 L at STP.
L H2 @ STP = mols H2 x 22.4 L/mol = ?.

how do yu se up the equation for this Q.??

You didn't ask a question.

I don't understand it either!!!

.500g

A.500

B.750
C.16.8

C'mon. Get serious. If I don't know what you don't understand about this I can't help you. To start can you balance the equation? If so do so. If not why can't you. Go from there with the next thing you don't understand.

Did you know?

Did you know that you can calculate the number of moles of a substance involved in a chemical reaction by using the balanced equation? In this case, when 13.5 grams of Al react with HCl, we can find the number of moles of Al by dividing its mass by its molar mass. To balance the equation, we need two molecules of HCl to react with two atoms of Al, resulting in two molecules of AlCl3 and three molecules of H2. So, a. 13.5 grams of Al is equal to 0.5 moles of Al reacting. And b. since the balanced equation shows that three moles of H2 are produced per two moles of Al, we can calculate the number of moles of H2 by multiplying the number of moles of Al by the ratio of H2 to Al. Therefore, 0.5 moles of Al produces 0.75 moles of H2. Lastly, c. we can use the ideal gas law to find the volume of H2 at STP. At STP, one mole of any gas occupies 22.4 liters. Thus, 0.75 moles of H2 will occupy 0.75 x 22.4 = 16.8 liters of space at STP.