What temperature must be maintained to ensure that a 1.00 L flask containing 0.0400 mol of oxygen will show a continuous pressure of 0.981 atm?

The temperature is 298.9K = 25.75 °C

Explanation:

Step 1: Data given

Volume of the flask = 1.00 L

Number of moles oxygen = 0.0400 moles

Pressure = 0.981 atm

Step 2: Calculate the temperature

p*V = n*R*T

⇒with p = the pressure of oxygen gas = 0.981 atm

⇒with V = the volume of the flask = 1.00 L

⇒with n = the number of moles of oxygen = 0.0400 moles

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = the temeprature = TO BE DETERMINED

T = (p*V) / (n*R)

T = (0.981 atm * 1.00 L) / (0.0400 moles * 0.08206 L*atm/mol*K)

T = 298.9 K

The temperature is 298.9K = 25.75 °C

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

the correct answer would be b) volume versus temperature, because they are also directly proportional.