Question

For the reaction $2Nal+C{{l}_{2}}\to 2NaCl+{{I}_{2}}$ . How many grams of $Nacl$ is attained if $5.0$ grams of $C{{l}_{2}}$ ?

Answer 1

First try to find the limiting reactant, first calculate the molar mass $NaI$ and $Cl$ then we have $5$ grams of $NaI$ and also $5$ grams of $C{{l}_{2}}$ and convert these amounts into moles.

Complete step by step answer:
We have the balanced equation $2NaI+C{{l}_{2}}\to 2NaCl+{{I}_{2}}$
let’s first find the molar mass
$NaI$ has a molar mass of $149.89g/mol$
$2NaI$ has a molar mass of $=2\times 149.89=299.78g/mol$
An average $Cl$ atom has a molar mass $=35.453g/mol$
So, the average molar mass of $C{{l}_{2}}=2\times 35.453=701.906g/mol$
The mole ratio between $NaI$ to $C{{l}_{2}}$ is $2:1$ , So two moles of sodium iodide or needed to react with one mole of chlorine.
$mo{{l}_{NaI}}=\dfrac{5g}{299.18g/mol}=1.67\times {{10}^{-2}}mol$
$mo{{l}_{C{{l}_{2}}}}=\dfrac{5g}{70.906g/mol}=7.05\times {{10}^{-2}}mol$
We have $7.05\times {{10}^{-2}}mol$ of $C{{l}_{2}}$ then
$7.05\times {{10}^{-2}}=1.47\times {{10}^{-1}}mol$ of $NaI$
We have $1.67\times {{10}^{-2}}mol$ of $NaI$ and $1.67\times {{10}^{-2}}<1.41.10$ so $NaI$ .
So, $1.67\times {{10}^{-2}}$ moles of sodium iodide will react there is no more of it.
The mole ratio between $NaI$ and $Nacl$ is $2:2=1$
So if $1.67\times {{10}^{-2}}$ mol of $NaI$ are used.
also $1.67\times {{10}^{-2}}$ mol of $Nacl$ are created.
Now, we need to find the amount in grams.
Mass $=$ moles $\times$ molar mass
The molar mass of $Nacl$ is $58.4g/mol$
${{m}_{Nacl}}=1.67\times {{10}^{-2}}mol\dfrac{58.4}{mol}$

**Note:** The coefficients in a balanced equation represent the no. of molecules or atoms that are reacting and are produced. For example- In the formation of water, $$2$$ molecules of hydrogen gas react with $$1$$ molecules of oxygen producing $$2$$ molecules of water.