Consider 2 al + 6 hcl → 2 alcl3 + 3 h2 , the reaction of al with hcl to produce hydrogen gas. what is the pressure of h2 if the hydrogen gas collected occupies 14.0 l at 300.k and was produced upon reaction of 4.50 moles of al and excess hcl in a process that has a 75.4 percent yield?

Answer : The pressure of hydrogen gas is 8.96 atm.

Explanation :

The given balanced chemical reaction is:

From the balanced chemical reaction we conclude that,

As, 2 moles of Al react to give 3 moles of gas

So, 4.50 moles of Al react to give moles of gas

Now we have to calculate the moles of gas when percent yield is 75.4.

Now we have to calculate the pressure of gas.

Using ideal gas equation :

PV = nRT

where,

P = Pressure of hydrogen gas = ?

V = Volume of the hydrogen gas = 14.0 L

n = number of moles of hydrogen gas = 5.09 moles

R = Gas constant =

T = Temperature of hydrogen gas = 300 K

Putting values in above equation, we get:

Therefore, the pressure of hydrogen gas is 8.96 atm.

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g