Chemistry help, Balance equation
February 1st, 2020
Okay.
First step is to balance all the ions, so like Al and Cl.
Al + 3HCl -> AlCl3 + H2
Alright, now we balance the H.
Al + 3HCl -> AlCl3 + 3/2H2
Or if you aren’t allowed fractional atoms, just times everything by 2
(because thats on the bottom of the fraction, sometimes this is more complicated if there are multiple fractions)
2Al + 6HCl -> 2AlCl3 + 3H2
Now count all the atoms to make sure
LHS:
2*Al
6*H
6*Cl
RHS:
2*Al
6*H
6*Cl
Done 😀
Need anymore help? Just post or PM me
Go on youtube and search for videos about balancing equations. They are pretty difficult first time doing it but after a couple of times, I am sure you will have no problems
This topic has been posted in the wrong section.
Moving.
Okay.
First step is to balance all the ions, so like Al and Cl.
Al + 3HCl -> AlCl3 + H2
Alright, now we balance the H.
Al + 3HCl -> AlCl3 + 3/2H2
Or if you aren't allowed fractional atoms, just times everything by 2
(because thats on the bottom of the fraction, sometimes this is more complicated if there are multiple fractions)
2Al + 6HCl -> 2AlCl3 + 3H2
Now count all the atoms to make sure
LHS:
2*Al
6*H
6*Cl
RHS:
2*Al
6*H
6*Cl
Done
Need anymore help? Just post or PM me
Thank You, I kind of get it now, but the true is that I don’t pay attention in class ahhah but Ima start paying attention, anyways the teacher gaves us a chart that says Formula, No, of moles Grams, Particles, and at the top it says Reactan 1, Reactant 2, Product 1, and Product 2, than she toll us to answers this question, I dont think I need help with the chart I just need help with the question,
1. How many moles of ALCL 3 will be produced when we use 12 moles of HCI?
2. how many grams of AL will be required to produce 108.12g of H2
3. If 9.03×10(at the top of the 10 its has does small numbers i forgot the name of it, those two numbers are 23) Particles of HCL are used, how many Particles of AL will be need for this reaction?
Thanks for all your help
1. How many moles of ALCL 3 will be produced when we use 12 moles of HCI?
2Al + 6HCl -> 2AlCl3 + 3H2
_____^^12 moles of this.
Okay, we already have 6 moles of HCl, so we just times the whole equation by 2 again.
4Al + 12HCl -> 4AlCl3 + 6H2.
_____________^^4 moles.
So the answer is 4.
2. how many grams of AL will be required to produce 108.12g of H2
Ok, first find out how many moles of H2 we are producing.
Mass of 1 mol H2 = 2.02g.
So moles of H2 = 108.12/2.02 = 53.5
the equation shows 3 moles, so we divide 53.5 by 3 to get 17.8.
(we dont have to, 17.8 is the number we are looking for)
Okay, there are two moles of Al on the left hand side.
times that by 17.8, and we get 35.7 moles.
Times this by the molar mass of Al, 26.98154 and we get our answer.
963.9grams.
(this seems wrong , I’ll check over it
icarbon seems to have a similar answer, so its probably right, you may wish to do it yourself, I have rounded my numbers in very weird places )
3. If 9.03×10^23 Particles of HCL are used, how many Particles of AL will be need for this reaction?
Okay, this is simlpe mostly the same logic, just converting into moles first…
okay, 1 mol is 6.022, so we divide 9.03 by this to get the number of moles
(Note I have am ignoring the ^23, they will get cancelled when we divide)
That gives us 1.5 moles.
2Al + 6HCl -> 2AlCl3 + 3H2
____^^1.5 moles of this.
So 1.5/6 gives us .25. This means that Al has 0.25 its original quantity.
2*0.25=0.5
Pretty simple now, just convert to #atoms by timesing with 6.022^23
3.011*10^23 atoms of Al
1. How many moles of ALCL 3 will be produced when we use 12 moles of HCI? 2Al + 6HCl -> 2AlCl3 + 3H2Thank you so much , this really helped me, Also Thank You icarbon
_____^^12 moles of this.
Okay, we already have 6 moles of HCl, so we just times the whole equation by 2 again.
4Al + 12HCl -> 4AlCl3 + 6H2.
_____________^^4 moles.
So the answer is 4.
2. how many grams of AL will be required to produce 108.12g of H2 Ok, first find out how many moles of H2 we are producing.
Mass of 1 mol H2 = 2.02g.
So moles of H2 = 108.12/2.02 = 53.5
the equation shows 3 moles, so we divide 53.5 by 3 to get 17.8.
(we dont have to, 17.8 is the number we are looking for)
Okay, there are two moles of Al on the left hand side.
times that by 17.8, and we get 35.7 moles.
Times this by the molar mass of Al, 26.98154 and we get our answer.
963.9grams.
(this seems wrong , I'll check over it
icarbon seems to have a similar answer, so its probably right, you may wish to do it yourself, I have rounded my numbers in very weird places )
that decimal point will bring a difference in the answers, your answers are more accurate (but both are right)for some one who dosnt pay attention in class, values on the periodic table is enough
Hahha come on I have that class 1st Period its like 7:30 AM im all sleepy and the teacher dont like me so I just dont pay attention and do other stuff, but Ima start paying attention now
Oh gosh, my chemisty teacher was a ~notgoingtosayit~
Didn’t do homework -> Clean tables.
I was in my last year of secondary school too…
Going to be taking a chemistry paper at University too, just because uni experiments are better
Oh gosh, my chemisty teacher was a ~notgoingtosayit~
Didn't do homework -> Clean tables.
I was in my last year of secondary school too...
Going to be taking a chemistry paper at University too, just because uni experiments are better Hey I think I got ur mistake on problem 2Okay, there are two moles of Al on the left hand side.
times that by 17.8, and we get 35.7 moles. 2 x 17.8 is 35.6 correct me if im wrong
Yeah like I said, bad rounding.
Ideally you do the whole thing on a calculator, so you keep the unrounded values right untill the end, where you can round it.
Yeah like I said, bad rounding.
Ideally you do the whole thing on a calculator, so you keep the unrounded values right untill the end, where you can round it.yeah Thank You icarbon and for your help if you need anything just PM me and i’ll see if I can help you. Finally done with my Homework ahhahaha