An equilibrium mixture contains 0.600 mol of each of the products (carbon dioxide and hydrogen gas) and 0.200 mol of each of the reactants (carbon monoxide and water vapor) in a 1.00-l container. this is the equation: co(g)+h2o(> < -- co2(g) + h2(g). how many moles of carbon dioxide would have to be added at constant temperature and volume to increase the amount of carbon monoxide to 0.300 mol?

Answer : The moles of added will be 1.12 mole.

Solution :  Given,

Moles of and at equilibrium = 0.200 mol

Moles of and at equilibrium = 0.600 mol

First we have to calculate the concentration of at equilibrium.

Now we have to calculate the value of equilibrium constant.

The given equilibrium reaction is,

The expression of will be,

Now we have to calculate the moles of added.

Let the moles of added is 'x'.

The given equilibrium reaction is,

Initially              0.200   0.200         0.600    0.600

Added moles       0           0                 x             0

Change             +0.1       +0.1             -0.1         -0.1

Final                   0.3         0.3          (0.5+x)        0.5

The expression of will be,

Therefore, the moles of added will be 1.12 mole.

answer is: empirical formula is tl₂o. if we use 100 grams of compound, tha:

m(tl) = 96.2%  · 100 g ÷ 100%.

m(tl) = 96.2 g.

n(tl) = m(tl)  ÷ m(tl). n(tl) = 96.2 g  ÷ 204.4 g/mol.

n(tl) = 0.47 mol. m(o) = 3.77 g. n(o) = m(o)  ÷ m(o). n(o) = 3.77 g  ÷ 16 g/mol. n(ag) = 0.235 mol. n(tl) . n(o) = 0.47 mol : 0.235 mol /0.235. n(tl) . n(o) = 2 : 1.

m(tl₂o) = 204.4 g/mol  · 2 + 16 g/mol = 424.8 g/mol.

m(compound) = 456. 8 g/mol - 424.8 g/mol = 32 g/mol ( 2 oxygens); tl₂o₃.