Question

Stoichiometry

Answer 1

@DLS

Answer 2

Hello @Callisto! Could you give me a hand?

Answer 3

A closed vessel is found to contain 192g of Mg and 96g of O2. This mixture is burnt.

1. What is the limiting reagent?
2. Find the weight of the reagent in excess.
3. How many mL of a 2M solution of H2SO4 will dissolve the residue in the vessel?

Answer 4

Is the chemical equation like this?
\[2Mg +O_2 \rightarrow 2MgO\]

Answer 5

Yes!

Answer 6

Have you learnt the concept of mole?

Answer 7

Yes :)

Answer 8

Ok.
\[\text{No. of moles} = \frac{\text{Mass}}{\text{Molar mass}}\]

No. of moles of Mg = 192/ 24.3 = ...
No. of moles of \(O_2\) = 96 / (16.0 *2) = ...
1 mole of \(O_2\) requires 2 moles of Mg for complete reaction. Then, you will know which one is in excess

Answer 9

In India, we just round off, so I guess it'd be OK to round off the molar mass...

192/24 = 8 moles of Mg
96/32 = 3 moles of O2

Oh, so 2g excess of Mg and O2 remains constant?

Answer 10

I mean O2 reacts completely, am I right?

Answer 11

Oh, that equivalent sign in my book finally makes sense

Answer 12

What exactly is the residue? :|

Answer 13

2 moles of Mg (sorry for writing grams back there) and 2 moles of MgO...

Answer 14

2 moles of MgO? Why?

Answer 15

Hmm... because for every 2 moles of Mg there are 2 moles of MgO?

Answer 16

So, the number of moles of Mg reacted is the same as the number of moles of MgO produced

Answer 17

192/24 = 8 moles of Mg
96/32 = 3 moles of O2

3 moles of O2 is reacted completely, so 2x3 = 6 moles of Mg is used.

Answer 18

6 reacted... 6 produced

Answer 19

Yup

Answer 20

Oh, sorry... got it

Answer 21

To my understanding, residue is the by-product. Though, I am not sure what it is in your question.

Answer 22

The thing in excess

Answer 23

^ not sure about it... just a guess

Answer 24

Do you have the answer?

Answer 25

Maybe I do, lemme check

Answer 26

Yeah... it does have the solved solution!

Answer 27

:O

Answer 28

I'm not able to understand it though

Answer 29

Is the residue 2 moles of Mg or...?

Answer 30

Yeah, it's exactly the same till what we did. It says, verbatim, "Now the residue i.g. Mg and MgO will react with H2SO4 to form sulphates"

Answer 31

i.e*

Answer 32

Where did that O come from?

Answer 33

Oh. All the remaining substances

\[Mg + H_2SO_4 \rightarrow MgSO_4 + H_2\]
\[MgO+H_2SO_4 \rightarrow MgSO_4+H_2O\]

Answer 34

Yeah, that is exactly what they gave! How is an O left though?

Answer 35

Which O are you talking about?

Answer 36

O as in MgO

Answer 37

Hmm... Have you learn that metal oxide + acid -> salt + water?

Answer 38

*learnt

Answer 39

Yes, but how was the oxygen atom in the residue? Didn't all the oxygen molecules get used up?

Answer 40

No lol
MgO <- this is formed after the oxidation of metal.
This compound reacts with H2SO4, not O.

Answer 41

Oh, that O added to Mg because of oxidation... OK...

Answer 42

Oh man... I was thinking all along that the residue is the thing in excess. That was the byproduct of the reaction. I got it!

Answer 43

Proceed, please :)

Answer 44

:/

Answer 45

Eh?

Answer 46

How many mL of a 2M solution of H2SO4 will dissolve the residue in the vessel?

Answer 47

Yes

Answer 48

Mg+H2SO4→MgSO4+H2
(i) You have 2 moles of Mg. How many moles of H2SO4 you need for complete reaction?

MgO+H2SO4→MgSO4+H2O
(ii) You have 6 moles of H2SO4. How many moles of H2SO4 you need for complete reaction?

(iii) Add the answer in (i) and (ii), you will get the total number of moles of H2SO4 you need for the complete reaction of residue.

Answer 49

Oh, right!

Answer 50

2 + 6 = 8

Answer 51

So 4 L of solution?

Answer 52

4000 mL

Answer 53

Thank you :)

Answer 54

You got it?

Answer 55

I need a better textbook and obviously, a better brain.

Answer 56

Yeah, thanks :)

Answer 57

You're welcome. FYI, I'm bad in Chemistry.

Answer 58

Hey, which texbook did you use?

Answer 59

Textbook

Answer 60

Yeah, but you can help me in 9th grade chemistry lol

Answer 61

http://www.scribd.com/doc/75639003/HKDSE-Chemistry-a-Modern-View-Book-1-Answers

Answer 62

Thanks, gonna check it out