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

The limiting reactant is the 6.279 g of
`NiCl_2`

Step-by-step 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~=~(mol)/(L)`

`0.1010~M~=(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(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:

`(0.0484)/(1)=0.0484`

`(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.