Carbon monoxide gas and hydrogen gas are mixed in a two liter flask. The reaction produces methanol gas as the only product. At equilibrium the Keq is 1.1 x 10-5. The flask initially held 6 grams of hydrogen and the equilibrium concentration of hydrogen is 0.1 M. How many grams of carbon monoxide were used?
mols H2 = 6/2 = 3
M H2 = 3mols/2L = 1.5M
.............CO + 2H2 ==> CH3OH
initial......?....1.5M.....0
change......-x.....-2x.....+x
equil......?-x......0.1.....x
H2 started at 1.5 and ended at 0.1; therefore, 2x = 0.7M
0.7M CH3OH produced. 0.7M CO used.
grams = mols x molar mass.
To find the number of grams of carbon monoxide used, we need to use stoichiometry and the given equilibrium concentration of hydrogen.
First, let's write the balanced equation for the reaction:
CO + 2H2 -> CH3OH
We are given that the equilibrium concentration of hydrogen is 0.1 M. Since the stoichiometric coefficient of hydrogen is 2, the equilibrium concentration of carbon monoxide would be 0.1 M / 2 = 0.05 M.
To find the number of moles of carbon monoxide used, we can use the formula: moles = concentration x volume.
The volume of the flask is 2 liters. So, the number of moles of carbon monoxide used is 0.05 M x 2 liters = 0.1 moles.
Now, we need to find the molar mass of carbon monoxide, which is approximately 28.01 g/mol.
Finally, we can calculate the number of grams of carbon monoxide used by multiplying the number of moles by the molar mass:
0.1 moles x 28.01 g/mol = 2.801 grams.
Therefore, approximately 2.801 grams of carbon monoxide were used.