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

2 Answers
Sep 8, 2016

Answer:

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?

Sep 8, 2016

Answer:

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.