A. Because you have to simplify
At STP conditions the volume of 1 mol of any ideal gas will be 22.4L
0.500 mol C3H3 x 22.4L / 1 mol = 11.2 L
<em>c</em> = 1.14 mol/L; <em>b</em> = 1.03 mol/kg
<em>Molar concentration
</em>
Assume you have 1 L solution.
Mass of solution = 1000 mL solution × (1.19 g solution/1 mL solution)
= 1190 g solution
Mass of NaHCO3 = 1190 g solution × (7.06 g NaHCO3/100 g solution)
= 84.01 g NaHCO3
Moles NaHCO3 = 84.01 g NaHCO3 × (1 mol NaHCO3/74.01 g NaHCO3)
= 1.14 mol NaHCO3
<em>c</em> = 1.14 mol/1 L = 1.14 mol/L
<em>Molal concentration</em>
Mass of water = 1190 g – 84.01 g = 1106 g = 1.106 kg
<em>b</em> = 1.14 mol/1.106 kg = 1.03 mol/kg
Answer : The amount of carbon dioxide produced is, 197.12 grams.
Explanation : Given,
Moles of ethanol = 2.24 mole
Molar mass of carbon dioxide = 44 g/mole
The balanced chemical reaction will be,

First we have to calculate the moles carbon dioxide.
From the balanced chemical reaction, we conclude that
As, 1 mole of ethanol react to give 2 moles of carbon dioxide
So, 2.24 mole of ethanol react to give
moles of carbon dioxide
Now we have to calculate the mass of carbon dioxide.


Therefore, the amount of carbon dioxide produced is 197.12 grams.