A gas mixture contains helium (He ), argon (Ar) , neon (Ne ), Xenon ( Xe), and krypton (Kr) gases. The total pressure of the gas mixture is 1252.5 mm Hg. The partial pressures of He, Ar , Ne, and Xe are 0.32, 0.21, 0.44, and 0.19 atm, respectively. What is the partial pressure exerted by Kr?

0.488atm = Pressure Kr

Explanation:

The total pressure in a mixture of gases could be defined as the sum of the partial pressures of a mixture. For the mixture of gases in the problem:

Total pressure = Pressure He + Pressure Ar + Pressure Ne + Pressure Xe + Pressure Kr

Converting the total pressure to atm:

1252.5mm Hg * (1atm / 760 mmHg) = 1.648 atm

Replacing:

1.648atm = 0.32atm + 0.21atm + 0.44atm + 0.19atm + Pressure Kr

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2