Question

Gaseous methane ch4 will react with gaseous oxygen o2 to produce gaseous carbon dioxide co2 and gaseous water h2o . suppose 1.44 g of methane is mixed with 9.5 g of oxygen. calculate the maximum mass of water that could be produced by the chemical reaction. round your answer to 3 significant digits.

Answer 1

Answer : The maximum mass of $H_2O$ produced will be,  3.24 grams

Explanation : Given,

Mass of $CH_4$ = 1.44 g

Mass of $O_2$ = 9.5 g

Molar mass of $CH_4$ = 16 g/mole

Molar mass of $O_2$ = 32 g/mole

Molar mass of $H_2O$ = 18 g/mole

First we have to calculate the moles of $CH_4$ and $O_2$.

$\text{Moles of }CH_4=\frac{\text{Mass of }CH_4}{\text{Molar mass of }CH_4}=\frac{1.44g}{16g/mole}=0.09moles$

$\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{9.5g}{32g/mole}=0.29moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CH_4+2O_2\rightarrow CO_2+2H_2O$

From the balanced reaction we conclude that

As, 1 moles of $CH_4$ react with 2 mole of $O_2$

So, 0.09 moles of $CH_4$ react with $2\times 0.09=0.18$ moles of $O_2$

From this we conclude that, $O_2$ is an excess reagent because the given moles are greater than the required moles and $CH_4$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2O$.

As, 1 mole of $CH_4$ react to give 2 moles of $H_2O$

So, 0.09 moles of $CH_4$ react to give $2\times 0.09=0.18$ moles of $H_2O$

Now we have to calculate the mass of $H_2O$.

$\text{Mass of }H_2O=\text{Moles of }H_2O\times \text{Molar mass of }H_2O$

$\text{Mass of }H_2O=(0.18mole)\times (18g/mole)=3.24g$

Therefore, the maximum mass of $H_2O$ produced will be, 3.24 grams.

Answer 2

The balanced equation for the above reaction is;
CH₄ + 2O₂ ---> CO₂ + 2H₂O
Stoichiometry of CH₄ to O₂ is 1:2
The number of methane moles present - 1.44 g/ 16 g/mol = 0.090 mol
Number of oxygen moles present - 9.5 g/ 32 g/mol = 0.30 mol
If methane is the limiting reagent,
0.090 moles of methane react with 0.090x 2 = 0.180 mol 
only 0.180 mol of O₂ is required but 0.30 mol of O₂ has been provided therefore O₂ is in excess and CH₄ is the limiting reactant.
Number of moles of water that can be produced - 0.180 mol
Therefore mass of water produced - 0.180 x 18 g/mol = 3.24 g 
Therefore mass of 3.24 g of water can be produced