Question

Calculate the no. of moles of $C{{O}_{2}}$ produced, if the initial amount of Carbon is $6g$ and oxygen is present in excess.
${{C}_{(s)}}+\dfrac{1}{2}{{O}_{2(g)}}\xrightarrow{{}}C{{O}_{(g)}}$ and $C{{O}_{(g)}}+\dfrac{1}{2}{{O}_{2(g)}}\xrightarrow{{}}C{{O}_{2(g)}}$

Answer 1

Hint : We know that in order to find the solution to this question, we have to multiply the given number of moles by Avogadro’s number. The mole is a unit of measurement for an amount of the substance in a system unit that is not standard but international.

Complete Step By Step Answer:
First thing you need to know is the molar mass of one mole of Oxygen and Carbon. So that, you can calculate the molar mass of the compound. A mole is a physical quantity which represents the amount of mass of the substance required to have a collective of atoms $6.0223\times {{10}^{23}}$ of the given substance.
Mole is widely used for units that calculate the amount of the matter of substance. One mole of every substance weighs about the same as the molecular mass of the similar substance.
Avogadro’s number of $6.0223\times {{10}^{23}}$ molecules per moles is used to calculate the number of molecules. It represents the number of atoms or molecules or entities present in one mole of a substance. The Avogadro’s law established how the gas amount (n) is related to its volume (v). A direct relationship is implied that the gas volume is completely proportional to the amount of gas moles present in it. Here we have to calculate the number of moles present in reactant and product.

Given Chemical Reaction:	Calculating moles of each compound, here we have taken $0.5moles\left( Carbon \right)$ so that addition of reactants is equal to no. of moles in the product which is here $1$ .	Equating moles and looking for Reactant moles $=$ Product moles, i.e. $L.H.S(1)=R.H.S(1)$
${{C}_{(s)}}+\dfrac{1}{2}{{O}_{2(g)}}\xrightarrow{{}}C{{O}_{(g)}}$	$0.5\left[ {{C}_{(s)}} \right]+\dfrac{1}{2}\left[ {{O}_{2(g)}} \right]\xrightarrow{{}}1\left[ C{{O}_{(g)}} \right]$	$\left( 0.5 \right)+\left( \dfrac{1}{2} \right)\to 1$ $\Rightarrow 0.5+0.5\to 1$
$C{{O}_{(g)}}+\dfrac{1}{2}{{O}_{2(g)}}\xrightarrow{{}}C{{O}_{2(g)}}$	$0.5\left[ C{{O}_{(s)}} \right]+\dfrac{1}{2}\left[ {{O}_{2(g)}} \right]\xrightarrow{{}}1\left[ C{{O}_{(g)}} \right]$	$\left( 0.5 \right)+\left( \dfrac{1}{2} \right)\to 1$ $\Rightarrow 0.5+0.5\to 1$

Note :
Remember that the mole concepts simplify mass relation among reactants and products like we can have base and our calculation on coefficient number of molecules involved in reaction. At the same time mass of substances are on a lab scale unit.