Question

45. Assume that 0.504 g of H₂ gas at STP reacts with

excess CuO according to the following equation:
Cuo(s) + H₂(g) → Cu(s) + H₂O(g)
Make sure the equation is balanced before beginning
your calculations.
a. How many liters of H₂ react?
b. How many moles of Cu are produced?
C. How many grams of Cu are produced?

Answer 1

The balanced equation for the reaction is:

CuO(s) + H2(g) → Cu(s) + H2O(g)

a. To find the volume of H2 gas at STP, we can use the ideal gas law:

PV = nRT

At STP, the pressure (P) is 1 atm and the temperature (T) is 273 K. The volume (V) of 1 mole of gas at STP is 22.4 L. So, the number of moles (n) of H2 gas is:

n = PV/RT = (1 atm) x V/(0.0821 L·atm/mol·K x 273 K) = V/22.4

We know that 0.504 g of H2 gas reacts. The molar mass of H2 is 2 g/mol. So, the number of moles of H2 gas is:

n = m/M = 0.504 g/2 g/mol = 0.252 mol

Now we can find the volume of H2 gas:

V = n x 22.4 L/mol = 0.252 mol x 22.4 L/mol = 5.65 L

Therefore, 5.65 liters of H2 gas react.

b. From the balanced equation, we can see that the mole ratio of CuO to Cu is 1:1. So, the number of moles of Cu produced is also 0.252 mol.

c. To find the mass of Cu produced, we can use the molar mass of Cu, which is 63.5 g/mol:

mass of Cu = number of moles of Cu x molar mass of Cu
= 0.252 mol x 63.5 g/mol
= 16.02 g

Therefore, 16.02 grams of Cu are produced.