Determine the temperature (in K) of 1.5 moles of gas contained in a 5.5 L vessel at a pressure of 192.4 kPa.

84.96K

Explanation:

The following were obtained from the question:

n = 1.5 mole

V = 5.5L

P = 192.4 kPa = 192400Pa

Recall

101325Pa = 1atm

192400Pa = 192400/101325 = 1.9atm

R = 0.082atm.L/K /mol

T=?

We can obtain the temperature by applying the following:

PV = nRT

T = PV / nR

T = (1.9 x 5.5)/(1.5 x 0.082)

T = 84.96K

what are the statements

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