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VCE Chemistry Question Thread

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lzxnl:

--- Quote from: eagles on January 04, 2014, 10:35:39 am ---When a certain non-metal whose formula is X8 burns in air XO3 forms. Write a balanced equation for this reaction. If 120.0g of oxygen gas is consumed completely, along with 80.0g of X8, identify element X.

Is this balanced equation correct: X8 (g) + 12 O2 (g) -> 8XO3 (g)
I'm not sure how to proceed with this question....

Your help is appreciated   :D

--- End quote ---

Sorry. The chemical formula makes it look slightly obvious as to what it is :P

Just saying, X8 looks like a solid to me. But you're not meant to know that.

Your equation is balanced correctly though. Now, you know that 120.0 g of oxygen gas is used up. This is 120/32=15/4 moles of oxygen gas (divide by 32 not 16 remember). This must have reacted with exactly 15/48 moles of X8 by mole ratios. If this mass of X8 is 80 grams, one mole of your unknown compound is 80*48/15=80*16/5=256 grams
This is for X8, so the molar mass of X is 32 grams
AKA sulfur


How many non-metals are there anyway that exist as X8 and form trioxides? :P

brightsky:
Sulfur fits the bill quite nicely. Let see if it is indeed sulfur...

n(O2) = m/M = 120.0/(2*16.0) = 120.0/32.0 = 3.75 mol
n(X8) = m/M = 80.0/M(X8) = 3.75/12 = 0.3125 mol (since the n(X8) : n(O2) = 1 : 12)
M(X8) = 80.0/0.3125 = 256 g/mol
M(X) = 256/8 = 32 g/mol

The element X is indeed sulfur.

Yacoubb:
X8 reacts completely with oxygen; a combustion reaction.

X8(s) + 12O2(g) ---> 8XO3(s)

Now that we have our balanced equation, we use stoichiometry.

We know that we've reacted 120.0g of O2 with 80g of X8.

n(O2): n(X8)
12: 1
120/32: n
Cross multiply
n(X8) = 0.3125 mol

We want to find the molecular mass of this metal, so.

0.3125 = 80 / mr
Mr(X8) = 256
Mr(X) = 256 / 8
Mr(X) = 32

Thus, by consulting our periodic table, we realise the metal is Sulphur.

DJA:

--- Quote from: lzxnl on January 04, 2014, 10:42:40 am ---Sorry. The chemical formula makes it look slightly obvious as to what it is :P

AKA sulfur

How many non-metals are there anyway that exist as X8 and form trioxides? :P

--- End quote ---

lol

Nliu--> Looks at the question...A few seconds later knows what the answer is.
Who bothers to calculate these days anyway?

eagles:
Haha thanks for all your help and different ways to approaching the question.

I have another question:

A 10.0 mL sample of HNO3 was diluted to a volume of 100.0 mL. 25 mL of the dilute solution was needed to neutralise 50.0 mL of a 0.60 M KOH solution. What was the concentration of the original nitric acid?

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