Answer: 5.70M
Explanation:
Molar mass of CO = 28.01 g/mol
Molar mass of H2 = 2.02 g/mol
Molar mass of CH3OH = 32.05 g/mol.
To determine the amount of each compound in the reaction mixture we use the formula.
Amount in mol = reacting mass/molar mass.
Inputing the given values we have,
26.6 g CO x (1 mol / 28.01 g ) = 0.9496608354 mol of CO.
To calculate the concentration of CO we use C=n/v, where n=amount and v= volume of CO.
Inputing the values in the formula
[CO] = 0.9496608354 mol CO / 5.23 L = 0.18158 M CO
Repeating thesame procedure for H
Amount of H=2.36 g H2 x ( 1 mol / 2.02 g ) = 1.168316832 mol of H2
Concentration of H2 in the mixture
[H2] = 1.168316832 mol H2 / 5.23 L = 0.223388 M of H2
Amount of CH3OH is determine similarly using rmass/molar mass
8.66 CH3OH x (1 mol / 32.05 g ) = 0.2702028081 mol of CH3OH
Concentration of
[CH3OH] = 0.2702028081 mol CH3OH/ 5.23 L = 0.051664 M CH3OH
Now equilibrium constant is determined by
Kc = [CH3OH] / [CO] [H2]^2
=0.051664/0.18158×0.223388×0.223388.
=5.70
