Question

In this reaction, how many grams of O2 are required to completely react with 110 grams of Al

Answer 1

98 g

Step-by-step explanation:

Start with the balanced equation. Then, make a little chart under the equation showing the information you have and need.

3O₂(g) + 4Al(s) → 2Al₂O₃(s)

m ? 110 g

M ___ ____

n ___ <= ____

"m" is for mass. "M" is for molar mass (some teachers use "MM"). "n" is for the number of moles.

To find the mass of oxygen:

1. Calculate the molar mass of oxygen (
   `M_{O_(2)`)
2. Calculate molar mass of aluminum (
   `M_(Al)`)
3. Use
   `M_(Al)` to find the moles of aluminum (
   `n_(Al)`)
4. With
   `n_(Al)`, use the mole ratio to find the moles of oxygen (
   `n_{O_(2)}`)
5. Use
   `n_{O_(2)}` and
   `M_{O_(2)}` to find the mass of oxygen (
   `m_{O_(2)}`)

To find molar mass, use the atomic mass on your periodic table. For each atom of an element, add on its atomic mass.

Molar mass of aluminum (one Al atom):

`M_(Al) = 26.982 g/mol`

Molar mass of oxygen (two O atoms):

`M_{O_(2)} = 16.000g/mol+16.000g/mol`

`= 32.000g/mol`

Update the chart:

3O₂(g) + 4Al(s) → 2Al₂O₃(s)

m ? 110 g

M 32.000 g/mol 26.982 g/mol

n ___ <= ____

Find the moles of aluminum

`n_(Al) = (110g)/(1) *(1mol)/(26.982g)` Multiply mass by molar mass to find moles.

`n_(Al) = (110)/(1) *(1mol)/(26.982)` The units "g" cancel out.

`n_(Al) = 4.0(7)mol` Keep one extra significant figure. (110 has 2 sig. figs.)

3O₂(g) + 4Al(s) → 2Al₂O₃(s)

m ? 110 g

M 32.000 g/mol 26.982 g/mol

n ___ <= 4.0(7) mol

Find the moles of oxygen using the mole ratio, which comes from the coefficients in the balanced equation.

The mole ratio of oxygen to aluminum is 3 to 4.

`n_{O_(2)} = (4.0(7)mol_(Al))/(1)*(3mol_(O2))/(4mol_(Al))` Multiply moles of aluminum by the mole ratio.

`n_{O_(2)} = (4.0(7))/(1)*(3mol_(O2))/(4)` "molAl" units cancel out.

`n_{O_(2)} = 3.0(52)mol_(O2)` Keep two sig. figs. when the first is a "5"

3O₂(g) + 4Al(s) → 2Al₂O₃(s)

m ? 110 g

M 32.000 g/mol 26.982 g/mol

n 3.0(52) mol <= 4.0(7) mol

Find the mass of oxygen

`m_{O_(2)} = (3.0(52)mol)/(1)*(32.000g)/(1mol)` Multiply moles by molar mass.

`m_{O_(2)} = (3.0(52))/(1)*(32.000g)/(1)` The "mol" unit cancels out.

`m_{O_(2)} = 97.(6)g` Keep one sig. fig. to round. "6" rounds up.

`m_{O_(2)} = 98g` <= Final answer

∴ 98 grams of oxygen are required to completely react with 110 grams of aluminum.