Question

If you started with 7.461 g of magnesium metal, how much oxygen would be consumed during combustion?

Answer 1

A shade under `5*g` of `O_2` gas.

Explanation:

`Mg(s)+1/2O_2(g) rarr MgO(g)`

`"Moles of metal"` `=` `(7.461*g)/(24.305*g*mol^-1)=0.3070*mol`

And thus `0.1535*mol` dioxygen gas would be consumed. What is the mass of this quantity of gas? Under standard laboratory conditions, what would be its volume?

Answer 2

The amount of oxygen gas consumed is 4.911 g.

Explanation:

Start with a balanced equation.

`"2Mg(s)" + "O"_2("g)"` `rarr` `"2MgO(s)"`

The coefficient in front of each formula is its number of moles. No coefficient is understood to be 1. The mole ratio between magnesium and oxygen gas is `"2 mol Mg"/"1 mol O"_2"` or `"1 mol O"_2/"2 mol Mg"`.

You will need the molar masses of magnesium and oxygen gas from the periodic table.

Magnesium: `"24.305 g/mol"`
Oxygen gas (dioxygen): `(2xx15.999 "g/mol")="31.998 g/mol"`

First determine the moles of Mg in 7.461 g by dividing the 7.461 g by its molar mass.

`(7.461 cancel"g Mg")/(24.30cancel("g")/"mol Mg")="0.3069736 mol Mg"`

Determine the moles `"O"_2"` by multiplying the moles Mg by the mole ratio with oxygen on top.

`0.3069736 cancel"mol Mg"xx"1 mol O"_2/cancel"2mol Mg"="0.1534868 mol O"_2"`

Determine the mass of `"O"_2"` by multiplying moles `"O"_2"` by its molar mass.

`0.1534868 cancel"mol O"_2xx"31.998 g O"_2/cancel"1 mol O"_2"="4.911 "g O"_2"` (rounded to four significant figures.