Question

Not only due trees "fix" carbon but so do green vegetables. Photosynthesis in spinach leaves produces glucose via the Calvin cycle which involves the fixation of CO2 with ribulose 1-5 bisphosphate to form 3-phosphoglycerate via C3H8P2011(aq) + H2O(aq) + CO2(g) → 2 CzH4PO3(aq) + 2 H+(aq) If 15.0 g of 3-phosphoglycerate is formed by this reaction at T = 298 K and P = 1.00 atm what volume of CO2 is fixed? [1.00 L]

Answer 1

Answer : The volume of $CO_2$ gas is 1.00 L

Explanation :

First we have to determine the moles of $C_3H_4PO_7^{3-}$.

Molar mass of $C_3H_4PO_7^{3-}$ = 182.9 g/mole

$\text{ Moles of }C_3H_4PO_7^{3-}=\frac{\text{ Mass of }C_3H_4PO_7^{3-}}{\text{ Molar mass of }C_3H_4PO_7^{3-}}=\frac{15.0g}{182.9g/mole}=0.0820moles$

Now we have to calculate the moles of $CO_2$.

The given balanced chemical reaction is:

$C_5H_8P_2O_{11}^{4-}(aq)+H_2O(aq)+CO_2(g)\rightarrow 2C_3H_4PO_7^{3-}(aq)+2H^+(aq)$

From the reaction we conclude that,

As, 2 moles of $C_3H_4PO_7^{3-}$ produce from 1 mole of $CO_2$

So, 0.0820 moles of $C_3H_4PO_7^{3-}$ produce from $\frac{0.0820}{2}=0.041moles$ of $CO_2$

Now we have to calculate the volume of $CO_2$ gas.

Using ideal gas equation:

$PV=nRT$

where,

P = pressure of gas = 1.00 atm

V = volume of gas = ?

T = temperature of gas = 298 K

n = number of moles of gas = 0.041 mole

R = gas constant = 0.0821 L.atm/mole.K

Now put all the given values in the ideal gas equation, we get:

$(1.00atm)\times V=(0.041mole)\times (0.0821L.atmK^{-1}mol^{-1})\times (298K)$

$V=1.00L$

Therefore, the volume of $CO_2$ gas is 1.00 L