Respuesta :
Answer:
The limiting reactant is the 6.279 g of [tex]NiCl_2[/tex]
Explanation:
We have to start with the reaction between sodium carbonate ([tex]Na_2CO_3[/tex]) and the Nickel (II) Chloride ([tex]NiCl_2[/tex]), so:
[tex]Na_2CO_3~+~NiCl_2-->~NiCO_3~+~NaCl[/tex]
We will have a double replacement reaction. Now we have to balance the reaction, so:
[tex]Na_2CO_3~+~NiCl_2-->~NiCO_3~+~2NaCl[/tex]
The next step is the calculation of the moles for each reactive. For [tex]Na_2CO_3[/tex] we have use the molarity equation:
[tex]M~=~\frac{mol}{L}[/tex]
[tex]0.1010~M~=\frac{mol}{0.5~L}[/tex]
[tex]mol~=~0.1010*0.5=~0.0505~mol~of~Na_2CO_3[/tex]
For the calculation of moles of [tex]NiCl_2[/tex] we have to use the molar mass of the compound (129.59 g/mol):
[tex]6.279~g~NiCl_2\frac{1~mol~NiCl_2}{129.59~g~NiCl_2}=~0.0484~mol~NiCl_2[/tex]
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:
[tex]\frac{0.0484}{1}=0.0484[/tex]
[tex]\frac{0.0505}{1}=0.0505[/tex]
The final step is to choose the smallest value. In this case is the value that correspond to [tex]NiCl_2[/tex]. Therefore [tex]NiCl_2[/tex] is the limiting reactive.
