The concentration of carbon monoxide (CO), a common air pollutant, is found in a room to be 5.7 x 10^-3 mg/cm^3. How many grams
1 answer:
The amount, in mg, of CO present in the room will be 191,520 mg.
<h3>Stoichiometric problem</h3>The concentration of the gas in the room is 5.7 x mg/cm3.
The dimension of the room is 3.5 m x 3.0 m x 3.2 m. This is equivalent to 350 cm x 300 cm x 320 cm.
We can obtain the volume of the room as:
350 x 300 x 320 = 33,600,000 cm3
The concentration is in mg/cm3, meaning that it is mass/volume.
Thus:
mass = concentration x volume = 5.7 x mg/cm3 x 33,600,000 cm3
= 191,520 mg
The mass of CO in the room is 191,520 mg
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Answer : The percent yield is, 83.51 %
Solution : Given,
Mass of Zn = 5.00 g
Mass of = 25.00 g
Molar mass of Zn = 65.38 g/mole
Molar mass of = 168.97 g/mole
Molar mass of Ag = 107.87 g/mole
First we have to calculate the moles of Zn and .
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, 0.1479 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, 0.1479 moles of react to give 0.1479 moles of
Now we have to calculate the mass of
Theoretical yield of = 15.95 g
Experimental yield of = 13.32 g
Now we have to calculate the percent yield.
Therefore, the percent yield is, 83.51 %