Question

If 495 milliliters of carbon dioxide at 25°C and 101.3 kilopascals reacts with excess water, what is the theoretical yield of carbonic acid?

Answer 1

Hes close its 1.27 g

Answer 2

Answer:- 1.24 g

Solution:- The balanced equation for the formation of carbonic acid by the reaction of carbon dioxide with water is:

$CO_2+H_2O\rightarrow H_2CO_3$

From balanced equation, there is 1:1 mol ratio between carbon dioxide and carbonic acid. So, moles of carbonic acid will be equal to the moles of carbon dioxide used.

Moles of carbon dioxide can be calculated using ideal gas law equation as it's volume, temperature and pressure are given.

we need to convert mL to L, degree C to kelvin and kilopascals to atm.

$495mL(\frac{1L}{1000mL})$  = 0.495 L

25 + 273 = 298 K

$101.3kPa(\frac{1atm}{101.3kPa)})$

= 1 atm

Ideal gas law equation is:

PV = nRT

We want to find out the n, so let's rearrange this:

$n=\frac{PV}{RT}$

R is the universal gas constant and it's value is $\frac{0.0821atm.L}{mol.K}$ .

Let's pug in the values in the equation and solve it for n.

$n=\frac{1*0.495}{0.0821*298}$

n = 0.02 mol

To convert the moles to grams we multiply the moles by the molar mass of carbonic acid.

Molar mass of carbonic acid = 2(1.008)+12.01+3(16.00)

= 2.016+12.01+48.00

= 62.03 gram per mol (rounded to two decimal places)

Let's multiply the moles by molar mass:

$0.02mol(\frac{62.03g}{mol})$

= 1.24 g

So, the theoretical yield of carbonic acid is 1.24 g.