The initial volume of a gas at a pressure of 3 atm is 5.2 L. What will the volume be if the pressure is increased to 15 atm?

1 L

Explanation:

Use Boyle's Law, which is defined as:

p1v1 = p2v2

p1 = 3atm

v1 = 5.2L

p2 = 15atm

v2 = ?

We are solving for v2 because all other variables are given.

p1v1 = p2v2

(3atm)(5.2L) = (15atm)(v2)

v2 = (3)(5.2) / (15) = 1.04 L ≈ 1 L

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