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

Sep 8, 2016

A shade under $5 \cdot g$ of ${O}_{2}$ gas.

Explanation:

$M g \left(s\right) + \frac{1}{2} {O}_{2} \left(g\right) \rightarrow M g O \left(g\right)$

$\text{Moles of metal}$ $=$ $\frac{7.461 \cdot g}{24.305 \cdot g \cdot m o {l}^{-} 1} = 0.3070 \cdot m o l$

And thus $0.1535 \cdot m o l$ dioxygen gas would be consumed. What is the mass of this quantity of gas? Under standard laboratory conditions, what would be its volume?

Sep 8, 2016

The amount of oxygen gas consumed is 4.911 g.

Explanation:

$\text{2Mg(s)" + "O"_2("g)}$$\rightarrow$$\text{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 $\text{2 mol Mg"/"1 mol O"_2}$ or $\text{1 mol O"_2/"2 mol Mg}$.

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

Magnesium: $\text{24.305 g/mol}$
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 $\text{O"_2}$ by multiplying the moles Mg by the mole ratio with oxygen on top.

$0.3069736 \cancel{\text{mol Mg"xx"1 mol O"_2/cancel"2mol Mg"="0.1534868 mol O"_2}}$

Determine the mass of $\text{O"_2}$ by multiplying moles $\text{O"_2}$ by its molar mass.

$0.1534868 \cancel{\text{mol O"_2xx"31.998 g O"_2/cancel"1 mol O"_2"="4.911 "g O"_2}}$ (rounded to four significant figures.