Respuesta :
Answer : The mass of calcium phosphate and the mass of sodium chloride that formed could be, 9.3 and 10.5 grams respectively.
Explanation : Given,
Mass of [tex]Na_3PO_4[/tex] = 12.00 g
Mass of [tex]CaCl_2[/tex] = 10.0 g
Molar mass of [tex]Na_3PO_4[/tex] = 164 g/mol
Molar mass of [tex]CaCl_2[/tex] = 111 g/mol
Molar mass of [tex]NaCl[/tex] = 58.5 g/mol
Molar mass of [tex]Ca_3(PO_4)_2[/tex] = 310 g/mol
First we have to calculate the moles of [tex]Na_3PO_4[/tex] and [tex]CaCl_2[/tex].
[tex]\text{Moles of }Na_3PO_4=\frac{\text{Given mass }Na_3PO_4}{\text{Molar mass }Na_3PO_4}[/tex]
[tex]\text{Moles of }Na_3PO_4=\frac{12.00g}{164g/mol}=0.0732mol[/tex]
and,
[tex]\text{Moles of }CaCl_2=\frac{\text{Given mass }CaCl_2}{\text{Molar mass }CaCl_2}[/tex]
[tex]\text{Moles of }CaCl_2=\frac{10.0g}{111g/mol}=0.0901mol[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is:
[tex]2Na_3PO_4+3CaCl_2\rightarrow 6NaCl+Ca_3(PO_4)_2[/tex]
From the balanced reaction we conclude that
As, 3 mole of [tex]CaCl_2[/tex] react with 2 mole of [tex]Na_3PO_4[/tex]
So, 0.0901 moles of [tex]CaCl_2[/tex] react with [tex]\frac{2}{3}\times 0.0901=0.0601[/tex] moles of [tex]Na_3PO_4[/tex]
From this we conclude that, [tex]Na_3PO_4[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]CaCl_2[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of [tex]NaCl[/tex] and [tex]Ca_3(PO_4)_2[/tex]
From the reaction, we conclude that
As, 3 mole of [tex]CaCl_2[/tex] react to give 6 mole of [tex]NaCl[/tex]
So, 0.0901 mole of [tex]CaCl_2[/tex] react to give [tex]\frac{6}{3}\times 0.0901=0.1802[/tex] mole of [tex]NaCl[/tex]
and,
As, 3 mole of [tex]CaCl_2[/tex] react to give 1 mole of [tex]Ca_3(PO_4)_2[/tex]
So, 0.0901 mole of [tex]CaCl_2[/tex] react to give [tex]\frac{1}{3}\times 0.0901=0.030[/tex] mole of [tex]Ca_3(PO_4)_2[/tex]
Now we have to calculate the mass of [tex]NaCl[/tex] and [tex]Ca_3(PO_4)_2[/tex]
[tex]\text{ Mass of }NaCl=\text{ Moles of }NaCl\times \text{ Molar mass of }NaCl[/tex]
[tex]\text{ Mass of }NaCl=(0.1802moles)\times (58.5g/mole)=10.5g[/tex]
and,
[tex]\text{ Mass of }Ca_3(PO_4)_2=\text{ Moles of }Ca_3(PO_4)_2\times \text{ Molar mass of }Ca_3(PO_4)_2[/tex]
[tex]\text{ Mass of }Ca_3(PO_4)_2=(0.030moles)\times (310g/mole)=9.3g[/tex]
Therefore, the mass of calcium phosphate and the mass of sodium chloride that formed could be, 9.3 and 10.5 grams respectively.
