Question

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

Answer 1

Answer:

98 g

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

$n_{Al} = \frac{110g}{1} *\frac{1mol}{26.982g}$    Multiply mass by molar mass to find moles.

$n_{Al} = \frac{110}{1} *\frac{1mol}{26.982}$       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.

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

$n_{O_{2}} = \frac{4.0(7)}{1}*\frac{3mol_{O2}}{4}$             "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

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

$m_{O_{2}} = \frac{3.0(52)}{1}*\frac{32.000g}{1}$             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.