Question

Determine the limiting reactant for the reaction of sodium carbonate and nickel(II) chloride using the quantities listed below. 6.279 g solid nickel(II) chloride 500.0 mL of 0.1010 M sodium carbonate

Answer 1

Answer:

The limiting reactant is the 6.279 g of $NiCl_2$

Explanation:

We have to start with the reaction between sodium carbonate ($Na_2CO_3$) and the Nickel (II) Chloride ($NiCl_2$), so:

$Na_2CO_3~+~NiCl_2-->~NiCO_3~+~NaCl$

We will have a double replacement reaction. Now we have to balance the reaction, so:

$Na_2CO_3~+~NiCl_2-->~NiCO_3~+~2NaCl$

The next step is the calculation of the moles for each reactive. For $Na_2CO_3$ we have use the molarity equation:

$M~=~\frac{mol}{L}$

$0.1010~M~=\frac{mol}{0.5~L}$

$mol~=~0.1010*0.5=~0.0505~mol~of~Na_2CO_3$

For the calculation of moles of $NiCl_2$ we have to use the molar mass of the compound (129.59 g/mol):

$6.279~g~NiCl_2\frac{1~mol~NiCl_2}{129.59~g~NiCl_2}=~0.0484~mol~NiCl_2$

The next step is the division of each mole value by the coefficient of each reactive of the balance reaction. In this case we have "1" for each reactive, so:

$\frac{0.0484}{1}=0.0484$

$\frac{0.0505}{1}=0.0505$

The final step is to choose the smallest value. In this case is the value that correspond to $NiCl_2$. Therefore $NiCl_2$ is the limiting reactive.