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

Hint: 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.