# 17.8 grams of CH_4 reacts with 55.4 g of O_2 to from CO_2 and H_20. What is the mass of H_20...

## Question:

17.8 grams of CH{eq}_4 {/eq} reacts with 55.4 g of O{eq}_2 {/eq} to from CO{eq}_2 {/eq} and H{eq}_2 {/eq}O. What is the mass of H{eq}_2 {/eq}O formed and how much excess reactant is leftover?

## Stoichiometry:

The limiting reactant is necessary to determine the theoretical yield of the product. When we are given two or more reactants and asked for the mass of product the first step is determining the limiting reactant. In the stoichiometry, we are doing the calculation regarding the mass and moles of either reactants or products.

We are given,

• {eq}Mass \ of \ CH_4 = 17.8 \ g {/eq}
• {eq}Mass \ of \ O_2 = 55.4 \ g {/eq}

We are given the mass of reactants from the methane gas combustion reaction and asked for the mass of water formed and mass of excess reactant leftover. To determine the mass of water, we need to first determine the limiting reactant. The reactant reacted completely and gives the maximum theoretical yield of the product called limiting reactant. The balanced reaction is necessary to determine the limiting reactant.

The balanced reaction :

{eq}CH_4 + 2 O_2 \rightarrow CO_2 + 2 H_2O {/eq}

Step 1) Moles of given reactants :

we know the formula,

{eq}Mole = \dfrac {Given \ mass }{Molar \ mass} {/eq}

By plugging the values,

{eq}\begin{align} Mole \ of \ CH_4 & = \dfrac { 17.8 \ g}{16.042 \frac {g}{mol}} = 1.11 \ moles \ of \ CH_4 \\ Mole \ of \ O_2 & = \dfrac { 55.4 \ g}{31.9988 \frac {g}{mol}} = 1.73 \ moles \ of \ O_2 \end{align} {/eq}

Step 2) Limiting reactant and moles of water :

Moles of water from the methane,

From the balanced reaction,

1 mole of {eq}CH_4 {/eq} = 2 moles of {eq}H_2O {/eq}

so, 1.11 moles of {eq}CH_4 {/eq} = x moles of {eq}H_2O {/eq}

{eq}\begin{align} x \ moles \ of \ H_2O & = 1.11 \ moles\ of\ CH_4 \times \dfrac{2 \ moles\ of\ H_2O} { 1 \ moles\ of\ CH_4} \\ & = 2.22 \ moles \ of \ H_2O \end{align} {/eq}

Moles of water from the oxygen,

From the balanced reaction,

2 mole of {eq}O_2 {/eq} = 2 moles of {eq}H_2O {/eq}

so, 1.73 moles of {eq}O_2 {/eq} = x moles of {eq}H_2O {/eq}

{eq}\begin{align} x \ moles \ of \ H_2O & = 1.73 \ moles\ of\ O_2 \times \dfrac{2 \ moles\ of\ H_2O} { 2 \ moles\ of \ O_2 } \\ & = 1.73 \ moles \ of \ H_2O \end{align} {/eq}

The moles of water is minimum from the reactant {eq}O_2 {/eq} than {eq}CH_4 {/eq}, hence limiting reactant is {eq}O_2 {/eq} and excess reactants is {eq}CH_4{/eq}

Step 3) Mass of water, {eq}H_2O {/eq} :

Moles of water from the above calculation is 1.73moles.

We know the formula for calculating the mass,

{eq}\begin{align} Mole & = \dfrac {Given \ mass }{Molar \ mass}\\ Mass & = Moles \times Molar \ mass \end{align} {/eq}

By plugging the values,

{eq}Mass \ of \ H_2O = 1.73 \ moles \times 18.015 \frac {g}{mol} = 31.2 \ g {/eq}

Step 4) Mass of excess reactant :

We determined that limiting reactant is oxygen, hence excess reactant is methane. First, we need to calculate how much excess moles of methane leftover after the reaction.

The moles of excess {eq}CH_4{/eq} after the reaction = Moles of {eq}CH_4{/eq} given - Moles of {eq}CH_4{/eq} used .

The moles of {eq}CH_4{/eq} used as follows :

From the balanced reaction,

2 mole of {eq}O_2 {/eq} = 1 moles of {eq}CH_4{/eq}

so, 1.73 moles of {eq}O_2 {/eq} = x moles of {eq}CH_4{/eq}

{eq}\begin{align} x \ moles \ of \ CH_4 & = 1.73 \ moles \ of \ O_2\times \dfrac{1 \ moles\ of\ CH_4 } { 2 \ moles\ of\ CH_4} \\ & = 0.866 \ moles \ of \ CH_4 \end{align} {/eq}

Moles of methane gas used in the reaction is 0.555 moles.

The moles of excess {eq}CH_4{/eq} after the reaction = 1.11 moles - 0.866 moles = 0.244 \ moles

Now we need to convert the mole to mass of methane using the formula,

{eq}\begin{align} Mass & = Moles \times Molar \ mass \\ Mass \ of \ CH_4 & = 0.244 \ moles \times 16.042 \frac {g}{mol} = 3.91 \ g \end{align} {/eq}

Thus, 31.2 grams of the mass of water is formed and 3.91 grams of excess reactant leftover.