Respuesta :

kusi5320
The equation tells you that 2 mol CO2 are used when 3 mol O2 are produced.
You produce 8.5g O21
Molar mass O2 = 32g/mol
mol O2 in 8.5g = 8.5/32 = 0.2656 mol O2
This will require 0.2656*2/3 = 0.1771 mol CO2 is used
Molar mass CO2 = 44g/mol
Mass of 0.1771 mol CO2 = 0.1771*44 = 7.792 g CO2 used
Answer correct to 2 significant figures = 7.8g CO2 used.

Answer : The mass of [tex]CO_2[/tex] used are 8.8 grams.

Solution : Given,

Mass of [tex]O_2[/tex] = 6.5 g

Molar mass of [tex]O_2[/tex] = 32 g/mole

Molar mass of [tex]CO_2[/tex] = 44 g

First we have to calculate the moles of [tex]O_2[/tex].

[tex]\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{6.5g}{32g/mole}=0.20moles[/tex]

Now we have to calculate the moles of [tex]CO_2[/tex]

The balanced chemical reaction is,

[tex]CO_2\rightarrow C+O_2[/tex]

From the balanced reaction we conclude that

As, 1 mole of [tex]O_2[/tex] produced from 1 mole of [tex]CO_2[/tex]

So, 0.20 mole of [tex]O_2[/tex] produced from 0.20 mole of [tex]CO_2[/tex]

Now we have to calculate the mass of [tex]CO_2[/tex]

[tex]\text{ Mass of }CO_2=\text{ Moles of }CO_2\times \text{ Molar mass of }CO_2[/tex]

[tex]\text{ Mass of }CO_2=(0.20moles)\times (44g/mole)=8.8g[/tex]

Therefore, the mass of [tex]CO_2[/tex] used are 8.8 grams.