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Given the reaction:

2 H2(g) + O2(g) → 2 H2O(l)


What is the total number of liters of O2(g) at STP needed to produce 6.0 × 1023 molecules of H2O(l)?

A) 33.6 L
B) 44.8 L
C) 11.2 L
D) 22.4 L


Sagot :

Neetoo

Answer:

V = 22.34 L

Explanation:

Given data:

Volume of O₂ needed = ?

Temperature and pressure = standard

Number of molecules of water produced = 6.0× 10²³

Solution:

Chemical equation:

2H₂ + O₂       →      2H₂O

Number of moles of water:

1 mole contain 6.022× 10²³ molecules

6.0× 10²³ molecules ×  1 mole  /  6.022× 10²³ molecules

0.99 mole

Now we will compare the moles of oxygen and water.

                  H₂O         :            O₂  

                    2            :              1

               0.996         :          0.996

Volume of oxygen needed:

PV = nRT

V = nRT/P

V = 0.996 mol × 0.0821 atm.L/mol.K ×  273.15 K / 1 atm

V = 22.34 L

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