A sample of coal has the following analysis (wt %). Moisture 1.1%, Fixed Carbon 74%, Volatile Matter 17.9%, Carbon 63.7%, Hydrog
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Answer : The mass of carbon monoxide form can be 2.8 grams.
Solution : Given,
Moles of C = 0.100 mole
Mass of = 8.00 g
Molar mass of = 32 g/mole
Molar mass of CO = 28 g/mole
First we have to calculate the moles of .
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
From the balanced reaction we conclude that
As, 2 mole of react with 1 mole of
So, 0.1 moles of react with
moles of
From this we conclude that, is an excess reagent because the given moles are greater than the required moles and
is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of
From the reaction, we conclude that
As, 2 mole of react to give 2 mole of
So, 0.1 moles of react to give 0.1 moles of
Now we have to calculate the mass of
Therefore, the mass of carbon monoxide form can be 2.8 grams.