Question

For the following unbalanced equation:
H2(g) + CI2(g) --> HCI(g)

How many grams of HCI can be produced from 18.6g of H2

Answer 1

So, 18.6 grams of H2 equates to 9.3 moles of H2 and 18.6 moles of H atoms. I think this means that you can make 18.6 moles of HCl, assuming you have enough Cl.
Now, since HCl has a molar mass of 36.46 g/mol, 18.6 moles of HCl would have a mass of 678.16 grams.

Answer 2

@Vocaloid can you explain this? I'm still alittle confused

Answer 3

the above method works, it's just a little roundabout

H2(g) + CI2(g) --> HCI(g)

personally i'd start by balancing the equation (notice that putting a 2 in front of HCl balances the equation)

H2(g) + CI2(g) --> 2HCI(g)

Answer 4

then it's just stoich

18.6g of H2 ---> divide by the molar mass of H2 ---> that gives you moles of H2 --> multiplying by 2 gives moles of HCl since 2 moles of HCl are produced for each 1 mole of H2 --> multiply by molar mass of HCl to get grams

Answer 5

2588

Answer 6

hm, nah, smokey's original calculation of 678-ish should be right

may I see your calculations? I may be able to pinpoint what went wrong

Answer 7

molar mass of Hcl is not 139-whatever, it's 36.46

Answer 8

oh okay

Answer 9

18.6g/2.0158 * 2 * 36.46 should give your your sol'n

Answer 10

672.84

Answer 11

good, they only give us 3 sig figs so round that to 673 g

Answer 12

This would be the same concept right

Answer 13

sort of

you need to calculate

a) how many moles of PCl3 you can make with 1.49 mol P4
and
b) how many moles of PCl3 you can make with 9.13 mol Cl2

then the smaller quantity from a) vs b) = your limiting reagent

Answer 14

pay attention to the mole ratios given by the equation

Answer 15

so I would divide for the first part

Answer 16

you need to set up a conversion factor between moles P4 and moles PCl3 based on the chemical equation

Answer 17

CI2 is limiting

Answer 18

I know that :S

Answer 19

Would i do P4/4PCi

Answer 20

you want to set up the conversion factor so that P4 cancels out.

Answer 21

ok

Answer 22

6?

Answer 23

6 moles of PCl? yeah (round to 5.96 as we are given three sig figs)

Answer 24

then repeat the process with Cl

Answer 25

54.8

Answer 26

6

Answer 27

to be more specific, 6.09 (remember, 3 sig figs)

Answer 28

so, you end up with less product using P4 so P4 is your limiting reagent, and from the calculations earlier, we know that 1.49 mol P4 makes 5.96 mol PCl3 ---> your solution