20 Liters of a mixture of N2 gas and CH4 gas was made up with a pressure of 700. Mm Hg and a temperature of 300. C. If the partial pressure of N2 gas was 250. Mm Hg, how many grams of CH4 were added to the mixture?

4.04g of CH₄ are in the mixture

Explanation:

In a mixture of gases, the total pressure is equal to partial pressure of each gas in the mixture.

In the problem, total pressure is 700mmHg and there are just 2 gases (N₂ ans CH₄) where the partial pressure of N₂ is 250mmHg. Thus, pressure of CH₄ is:

700mm Hg - 250mm Hg = 450mmHg. In atm:

450 mmHg ₓ (1 atm / 760mm Hg) = 0.5921 atm

Using PV = nRT

Where P is pressure (0.5921atm), V is volume (20L), R is gas constant (0.082atmL/molK), T is absolute temperature (300°C + 273.15 = 573.15K), it is possible to obtain moles -n- of the gas, thus:

PV / RT = n

0.5921atm×20L / 0.082atmL/molK×573.15K = n

0.252 moles CH₄ = n

As molar mass of CH₄ is 16.04 g/mol:

0.252 moles CH₄ × (16.04g / 1mol) = 4.04g of CH₄ are in the mixture

what does the table look like

> b takes place in the gas phase in an isothermal tubular reactor. the feed is one mole of a per one mole of c, an inert. the entering temperature and pressure are 427°c and 10 atm, respectively. the gas constant r = 0.08206 atm · l/mol · k.