Identify the correct set-up. How many moles of aluminum oxide would form if 25.0 g of aluminum react completely given the following chemical reaction: 4Al + 3O₂ → 2Al₂O₃?

Option 1. 25g Al x 1mol Al/26.982g Al x 2mol Al₂O₃/ 4mol Al

Explanation:

The equation for the reaction is given below:

4Al + 3O₂ → 2Al₂O₃

Molar Mass of Al = 26.982g/mol

Mass of Al = 25g

The number of mole in 25g of Al can be expressed as:

25g of Al = 25g x 1 mol / 26.982g

From the balanced equation above,

4 mol of Al produced 2 moles of Al₂O₃.

Therefore, 25g of Al (25g x 1 mol / 26.982g) will produce =

25g Al x 1mol Al/26.982g Al x 2mol Al₂O₃/ 4mol Al

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