Question

A mixture of \(0.160\) moles of \(\mathrm{C}\) is reacted with \(0.117\) moles of \(\mathrm{O}_{2}\) in a sealed, \(10.0\) L vessel at \(500.0 \mathrm{~K}\), producing a mixture of \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\)

\(3 C(t)+20_{2}(g) \longrightarrow 2 \operatorname{co}(g)+\operatorname{co}_{2}(g)\)

The total pressure is \(0.699\) atm. What is the partial pressure of \(C 0^{2}\)

Answer 1

Hope you will like my answer .

3C(s) + 20, 19) — 9 CO(g) + CO2(8) from stochiometry olar ratio of cand Og = 3:2 105:1 present Ratio of C and = = Og actually 0.1600-117 1.367 : 1 . c is limiting reagent. in moles of co produced = 2 x 0.160 = 0.1067 mol moles of coa produced = 1/3 x 0-160 mol to = 0.0533 mol. moles of O₂ consumed = moles of co produced 0.1067 mol - so = . Moles of Og left = 10.117 - D. 1067) mol 5 0.0103 mol moles of co 0.1067 i Xco - Total moles of 10.1067 + 0.0533 +0.0103) Partial Pressure = Total x mole fraction = 0.626 Pressure = 0.699 x 0.626 = 10:438/atm gas :

Answer 2

0.438 atm