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Given :
Mass of oxygen containing carbon monoxide (CO) is 2.666 gram .
To Find :
How many grams of carbon (C) would be present in carbon monoxide (CO) that contains 2.666 grams of oxygen (O) .
Solution :
By law of constant composition , a given chemical compound always contains its component elements in fixed ratio (by mass) and does not depend on its source and method of preparation.
So , volume of solution does not matter .
Moles of oxygen , .
Now , molecule of CO contains 1 mole of C .
So , moles of C is also 0.167 mole .
Mass of carbon , .
Therefore , mass of carbon is 2 grams .
Hence , this is the required solution .
Answer: The percent yield of this combination reaction is 41.3 %
Explanation : Given,
Mass of = 5.0 g
Mass of = 10.0 g
Molar mass of = 23 g/mol
Molar mass of = 71 g/mol
First we have to calculate the moles of and
.
and,
Now we have to calculate the limiting and excess reagent.
The balanced chemical equation will be:
From the balanced reaction we conclude that
As, 2 mole of react with 1 mole of
So, 0.217 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, 0.217 mole of react to give 0.217 mole of
Now we have to calculate the mass of
Molar mass of = 58.5 g/mole
Now we have to calculate the percent yield of this reaction.
Percent yield =
Actual yield = 5.24 g
Theoretical yield = 12.7 g
Percent yield =
Percent yield = 41.3 %
Therefore, the percent yield of this combination reaction is 41.3 %