Question

the volume of gas formed at STP when 5gram of CH4 is reacted with 16g of O2 is?

Answer 1

Answer:

5.6 L

Explanation:

First of all, write the balanced chemical equation for the given reaction as-

$CH_4 + 2O_2$ ⇒$CO_2 + 2H_2O$

From the above reaction wee can see that -

Moles of $CO_2$ produced by  1 mole of $CH_4$ = 1

"            " $CO_2$        "         "      1    "         $O_2$    = 0.5

Hence  $O_2$ is the limiting reagent

Now 1 mole of  $O_2$ = 0.5  moles of $CO_2$  

or 32 g of $O_2$ = 22 g of  $CO_2$  

∴ 16g of $O_2$ = 11g of  $CO_2$

⇒ No. of moles of  $CO_2$ formed = 11/44= 0.25 moles.

Molar volume of  $CO_2$ at STP = 22.4 L

∴ Volume of  $CO_2$ formed = 22.4 × 0.25 = 5.6 L

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