Stoichiometry

Question

What volume of hydrogen gas (at STP) contains the same number of hydrogen atoms as 1.00 g of glucose (C6H12O6)?

Answer 1

I think a much simpler way is as follows:

1 g glucose x 1 mole glucose/186 g = 0.00556 moles glucose

0.00556 moles glucose x 12 moles H/mole glucose = 0.0667 moles H (hydrogen, not hydrogen gas)

It takes 2 moles H to make 1 mole H2 gas, thus moles H2 gas = 0.0667 moles/2 = 0.0333 moles H2 gas

Since it is at STP, no need to use ideal gas law of PV = nRT since we know that 1 moles any gas = 22.4 liters.

Thus, 0.0333 moles H2 x 22.4 L/mole = 0.746 liters of H2 gas

Answer 2

First, find the number of hydrogen atoms in 1.00 g of glucose.

 

(1.00 g C₆H₁₂O₆) × (1 mol C₆H₁₂O₆ / 180.1 g C₆H₁₂O₆) × (12 mol H / 1 mol C₆H₁₂O₆) × (6.022×10²³ atoms / 1 mol H) = 4.012×10²² hydrogen atoms per gram of glucose

 

Now, use the Ideal Gas Law (PV=nRT) to calculate the volume of hydrogen gas needed.

At STP: P = 1 atm, T = 273.15 K

R = 0.08206 L·atm/(mol·K)

n = (4.012×10²² H atoms) / (6.022×10²³ atoms/mol) = 0.0666 mol H = 0.0333 mol H₂ gas

 

PV = nRT ⇒ V = nRT/P

V = (0.0333 mol H₂)(0.08206 L·atm/(mol·K))(273.15K) / (1 atm) = 0.746 L of hydrogen gas