Chemistry, 21.06.2020 20:57 aghamuzahirali4392

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

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Answer from: luceridiaaz

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:

Calculate the molar mass of oxygen ( $M_{O_{2}$ )Calculate molar mass of aluminum ( $M_{Al}$ )Use $M_{Al}$ to find the moles of aluminum ( $n_{Al}$ )With $n_{Al}$ , use the mole ratio to find the moles of oxygen ( $n_{O_{2}}$ )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.

Answer from: Quest

i believe the answer is 5,000

Answer from: Quest

Ithink the answer is b.

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