0.80 mol of CO(g) is passed into a 4.0 L rigid vessel containing excess FeO(s) and Fe(s) at a fixed temperature, where K_c = 0.65 for FeO(s) + CO(g) Fe(s) + CO2(g). Using M(Fe) = 56 g mol^-1, calculate the mass of iron in g formed by the time equilibrium is reached, correct to two decimal places.

JEE Advanced levelNumerical answerDifficulty 3 of 3ce102

0.80 mol of $CO (g)$ is passed into a 4.0 L rigid vessel containing excess $FeO (s)$ and $Fe (s)$ at a fixed temperature, where $K_{c} = 0.65$ for $FeO (s) + CO (g) Fe (s) + CO X_{2} (g)$ . Using $M (Fe) = 56 g mo l^{- 1}$ , calculate the mass of iron in g formed by the time equilibrium is reached, correct to two decimal places.

Numerical answer type, answer in g, tolerance 0.01.

Let

x

mol of

CO X_{2}

form. Then

K_{c} = [CO X_{2}] / [CO] = x / (0.80 - x)

(the volume cancels because

Δ n = 0

). So

x = 0.80 \times 0.65/1.65 = 0.3152

mol. Each mole of

CO X_{2}

produces one mole of

Fe

, so mass

= 0.3152 \times 56 = 17.648