Respuesta :

2 O2 + CH4 CO2 + 2 H2O 

What mass of CH4 is required to completely react with 100 grams of O2? 

mass CH4 = 100 g O2 x (1 mol O2 / 32 g) x (1 mol CH4 / 2 mol O2) x (16.05 g / 1 mol CH4) 
 25 grams CH4 

Answer : The mass methane required is, 25 g

Explanation : Given,

Mass of oxygen gas = 100 g

Molar mass of oxygen gas = 32 g/mole

Molar mass of methane gas = 16 g/mole

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{100g}{32g/mole}=3.125moles[/tex]

Now we have to calculate the moles of [tex]CH_4[/tex].

The balanced chemical reaction is,

[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]

From the balanced reaction we conclude that

As, 2 moles of [tex]O_2[/tex] react with 1 mole of [tex]CH_4[/tex]

So, 3.125 moles of [tex]O_2[/tex] react with [tex]\frac{3.125}{2}=1.563[/tex] moles of [tex]CH_4[/tex]

Now we have to calculate the mass of [tex]CH_4[/tex]

[tex]\text{Mass of }CH_4=\text{Moles of }CH_4\times \text{Molar mass of }CH_4[/tex]

[tex]\text{Mass of }CH_4=(1.563mole)\times (16g/mole)=25.008g=25g[/tex]

Therefore, the mass methane required is, 25 g

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