Question

How many grams of calcium carbide are produced when $4.0$ moles of carbon react with an excess of calcium carbonate ($CaC{O_3}$)?

Answer 1

Firstly, we need to adjust the condition of calcium carbide responding with an overabundance of calcium carbonate. At that point we can locate the mass which is delivered by $4.0$ moles of carbon.
Utilize the equilibrium condition:
$2{\text{CaC}}{{\text{O}}_3}\left( s \right) + 5{\text{C}}\left( {{\text{gr}}} \right)\mathop \to \limits^{\Delta \;\:\;\:} 2{\text{Ca}}{{\text{C}}_2}\left( s \right) + 3{\text{C}}{{\text{O}}_2}\left( g \right)$

Complete step by step answer:
I will deliberately utilize a strategy that makes you consider extensive and intensive properties...
The unbalanced condition was:
${\text{CaC}}{{\text{O}}_3}\left( s \right) + {\text{C}}\left( {{\text{gr}}} \right)\mathop \to \limits^{\Delta \;\:\;\:} {\text{Ca}}{{\text{C}}_2}\left( s \right) + {\text{C}}{{\text{O}}_2}\left( g \right)$
a) Balance the response. I picked calcium first:
$2{\text{CaC}}{{\text{O}}_3}\left( s \right) + {\text{C}}\left( {{\text{gr}}} \right)\mathop \to \limits^{\Delta \;\:\;\:} 2{\text{Ca}}{{\text{C}}_2}\left( s \right) + {\text{C}}{{\text{O}}_2}\left( g \right)$
At that point I picked oxygen:
$2{\text{CaC}}{{\text{O}}_3}\left( s \right) + {\text{C}}\left( {{\text{gr}}} \right)\mathop \to \limits^{\Delta \;\:\;\:} 2{\text{Ca}}{{\text{C}}_2}\left( s \right) + 3{\text{C}}{{\text{O}}_2}\left( g \right)$
At that point I picked carbon toward the end since $C$ is not difficult to adjust when a solitary atom is on one side to utilize.
$2{\text{CaC}}{{\text{O}}_3}\left( s \right) + 5{\text{C}}\left( {{\text{gr}}} \right)\mathop \to \limits^{\Delta \;\:\;\:} 2{\text{Ca}}{{\text{C}}_2}\left( s \right) + 3{\text{C}}{{\text{O}}_2}\left( g \right)$
We are informed that $\;CaC{O_3}$ is in abundance, so unmistakably, $C$ is the restricting reactant. Accordingly, since $5$ mols $C$ structure $2$ mols $Ca{C_2}$, we downsize the response to get:
$\dfrac{4}{5} \times \left( {2{\text{CaC}}{{\text{O}}_3}\left( s \right) + 5{\text{C}}\left( {{\text{gr}}} \right)\mathop \to \limits^{\Delta \;\:\;\:} 2{\text{Ca}}{{\text{C}}_2}\left( s \right) + 3{\text{C}}{{\text{O}}_2}\left( g \right)} \right)$
Presently we read straightforwardly from the re-scaled adjusted response that:
$\Rightarrow \dfrac{4}{5} \times {\text{5 moles C}}$ forms $\dfrac{4}{5} \times 2 = \dfrac{8}{5}{\text{mols Ca}}{{\text{C}}_2}$.
So, the required answer is $\left[ {\dfrac{8}{5}{\text{mols Ca}}{{\text{C}}_2}} \right]$
On the off chance that the molar mass of $Ca{C_2}$ is $64.099{\text{ }}g/mol$,
Is the number of grams of $Ca{C_2}$ includes $85mols$ of it? Your answer should be bigger than $64.099{\text{ }}g$ to make sense.
Calcium carbonate is a synthetic inorganic compound having the substance recipe $CaC{O_3}$. They are likewise called calcite. The image of calcium carbonate can be given as $CaC{O_3}$.
Calcium carbide is additionally called calcium acetylide, which is a synthetic compound having the substance recipe of $Ca{C_2}$. Basically, it is utilized mechanically for the creation of calcium cyanamide and acetylene.

So, the required answer is $\left[ {\dfrac{8}{5}{\text{mols Ca}}{{\text{C}}_2}} \right]$

Note: The calcium carbide applications incorporate acetylene gas producing and for the age of acetylene in carbide lights; synthetic compounds fabricating for manure; and furthermore, in steelmaking.
Calcium carbonate is likewise perhaps the most famous synthetic substance which is experienced first in school homerooms, where the utilization of chalk (which is a type of $CaC{O_3}$) is found.