An ideal gas had a mass of 0.0218g and occupied a volume of 1.111L at 0oC and 0.0100atm. What is the molar mass of this gas?

Explanation:

we shall use gas law formula to solve the problem .

P = .01 atm

V = 1.111 L

T = 273 K

mass of gas = .0218

molar mass = M

no of moles n = .0218 / M

PV = n RT

.01 x 1.111 = (.0218 / M)  x .082 x 273

.0218 / M = .000496

M = .0218 / .000496

= 43.95

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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