2C\(_2\)H\(_2\)(g) + 5O\(_2\)(g) → 4CO\(_2\)(g) + 2H\(_2\)O(g)
In the reaction above, the mass of carbon(IV)oxide produced on burning 78 g of ethyne is
[C =12, O = 16, H = 1]

264 g

39 g

352 g

156 g

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Vincere

12 years ago

according 2 d rxn,52g of ethyne require 176g of CO2.

52g_____________176g

78g_____________176/52 X 78/1

3.385g X 78=264g.

therefore,

78g of ethyne require 264g of carbon(iv) oxide.so d ansa is A

chydirim@gmail.com

10 years ago

Here is the correct explanation:

2 moles of C2H2- 4 moles of CO2

2*26- 4*44

52g of ethyne will give 176g of carbon(iv)oxide

78g of ethyne will give xg of carbon(iv) oxide

52/78=176/x

176*78=52x

13728=52x

x=13728/52

x=264. Option A

lilrichy

11 years ago

The answer is A not C..

no. of moles of Ethyne is 3mol

2mole of ethyne = 4mole of carbon(iv)oxide

therefore no of mole of carbon(iv)oxide is

(3*4)/2 = 6mol

mass of carbon(iv)oxide = 6 * (44)

= 264g of Carbon(iv)oxide

Myschool Amos

10 years ago

The question has been modified.

DrUche

11 years ago

Much has already been said but I still want to add that Option A was just right.

Sammy boi

12 years ago

Yea dey represented the ratio well bt there was a miss calculation along so d ans is A

mickael

10 years ago

The selected answer is wrong:

2moles of C2H2 produce 4moles of CO2,it therefore implies that 2*26g of C2H2 produces 4*44g of CO2 52g produces 176g 78g produces xg of CO2;cross multiply 78*176/52=264g