If a reaction vessel initially contains 5 mol S and 10 mil O2, how many moles of S will be in the reaction vessel once the reactants have reacted as much as possible

Explanation:

2S       +       3O₂       =       2SO₃

2moles      3 moles

2 moles of S react with 3 moles of O₂

5 moles of S will react with 3 x 5 / 2 moles of O₂

= 7.5 moles of O₂ .

O₂ remaining unreacted = 10 - 7.5 = 2.5 moles .

All the moles of S will exhausted in the reaction and 2.5 moles of oxygen will be left .  

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