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3 years ago

What are the coeffients that balance the following skeletal equation? CH4 + O2 + Cl2 ® HCI + CO

Chemistry

1 answer:

oee [108]3 years ago

8 0

Answer:

CH4 ---> 2

CL2 ----> 4

HCL -----> 8

CO-------> 2

Explanation:

Equation

CH4 + O2 + Cl2 => HCI + CO

we have to balance it

On LHS atoms of oxygen is 2 while on RHS it is 1 atom so lets balance oxygen first

to balance oxygen on RHS we add 2 as coefficient to CO

CH4 + O2 + Cl2 => HCI + 2CO

___________________________________________________

On RHS atoms of Carbon is 2 while on RHS it is 1 atom so lets balance Carbon now

to balance Carbon on LHS we add 2 as coefficient to CH4

2CH4 + O2 + Cl2 => HCI + 2CO

___________________________________

On LHS atoms of hydrogen is 8 so lets balance hydrogen now

to balance hydrogen on RHS we add 8 as coefficient to HCL

2CH4 + O2 + Cl2 => 8HCI + 2CO

______________________________________________

On RHS atoms of Cl is 8 while on LHS atoms of Cl is 2.

To make no of atoms of Cl 8 on LHS we add 4 as coefficient

2CH4 + O2 + 4Cl2 => 8HCI + 2CO

Thus, coefficients are as given below

CH4 ---> 2

CL2 ----> 4

HCL -----> 8

CO-------> 2

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<u>Answer:</u> The volume of hydrogen gas collected over water is 2.13 L

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To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of zinc = 5.566 g

Molar mass of zinc = 65.4 g/mol

Putting values in above equation, we get:

$\text{Moles of zinc}=\frac{5.566g}{65.4g/mol}=0.0851mol$

For the given chemical reaction:

$Zn(s)+2HCl(aq.)\rightarrow H_2(g)+ZnCl_2(aq.)$

As, HCl is present in excess. So, it is considered as an excess reagent.

Zinc is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of zinc produces 1 mole of hydrogen gas.

So, 0.0851 moles of zinc will produce = $\frac{1}[1}\times 0.0851=0.0851mol$ of hydrogen gas

To calculate the volume of hydrogen gas, we use ideal gas equation, which is:

PV = nRT

where,

P = Pressure of hydrogen gas = Total atmospheric pressure - vapor pressure of water = (752 - 18.65) mmHg = 733.35 mmHg

V = Volume of the hydrogen gas

n = number of moles of gas = 0.0851 moles

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

T = Temperature of hydrogen gas = $21^oC=[21+273]K=294K$

Putting values in above equation, we get:

$733.35mmHg\times V=0.0851mol\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 294K\\\\V=\frac{0.0851\times 62.3637\times 294}{733.35}=2.13L$

Hence, the volume of hydrogen gas collected over water is 2.13 L

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