Respuesta :
Answer: The limiting reagent is iron (II) chloride and the mass of sodium chloride formed is 21.2 grams
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
- For iron (II) chloride:
Given mass of iron (II) chloride = 23 g
Molar mass of iron (II) chloride = 126.8 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of iron (II) chloride}=\frac{23g}{126.8g/mol}=0.181mol[/tex]
- For sodium phosphate:
Given mass of sodium phosphate = 41 g
Molar mass of sodium phosphate = 164 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of sodium phosphate}=\frac{41g}{164g/mol}=0.25mol[/tex]
The chemical equation for the reaction of iron (II) chloride and sodium phosphate follows:
[tex]3FeCl_2+2Na_3PO_4\rightarrow 6NaCl+Fe_3(PO_4)_2[/tex]
By Stoichiometry of the reaction:
3 moles of iron (II) chloride reacts with 2 moles of sodium phosphate
So, 0.181 moles of iron (II) chloride will react with = [tex]\frac{2}{3}\times 0.181=0.1206mol[/tex] of sodium phosphate
As, given amount of sodium phosphate is more than the required amount. So, it is considered as an excess reagent.
Thus, iron (II) chloride is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
3 moles of iron (II) chloride produces 6 mole of sodium chloride.
So, 0.181 moles of iron (II) chloride will produce = [tex]\frac{6}{3}\times 0.181=0.362moles[/tex] of sodium chloride.
Now, calculating the mass of sodium chloride from equation 1, we get:
Molar mass of sodium chloride = 58.5 g/mol
Moles of sodium chloride = 0.362 moles
Putting values in equation 1, we get:
[tex]0.362mol=\frac{\text{Mass of sodium chloride}}{58.5g/mol}\\\\\text{Mass of sodium chloride}=(0.362mol\times 58.5g/mol)=21.2g[/tex]
Hence, the limiting reagent is iron (II) chloride and the mass of sodium chloride formed is 21.2 grams
