Answer : The mass of
produced are 5.3 grams.
Solution : Given,
Mass of
= 10.0 g
Mass of
= 10.0 g
Molar mass of
= 95.21 g/mole
Molar mass of
= 163.94 g/mole
Molar mass of
= 262.87 g/mole
First we have to calculate the moles of
and
.
![\text{ Moles of }MgCl_2=\frac{\text{ Mass of }MgCl_2}{\text{ Molar mass of }MgCl_2}=\frac{10.0g}{95.21g/mole}=0.105moles](https://tex.z-dn.net/?f=%5Ctext%7B%20Moles%20of%20%7DMgCl_2%3D%5Cfrac%7B%5Ctext%7B%20Mass%20of%20%7DMgCl_2%7D%7B%5Ctext%7B%20Molar%20mass%20of%20%7DMgCl_2%7D%3D%5Cfrac%7B10.0g%7D%7B95.21g%2Fmole%7D%3D0.105moles)
![\text{ Moles of }Na_3PO_4=\frac{\text{ Mass of }Na_3PO_4}{\text{ Molar mass of }Na_3PO_4}=\frac{10.0g}{163.94g/mole}=0.060moles](https://tex.z-dn.net/?f=%5Ctext%7B%20Moles%20of%20%7DNa_3PO_4%3D%5Cfrac%7B%5Ctext%7B%20Mass%20of%20%7DNa_3PO_4%7D%7B%5Ctext%7B%20Molar%20mass%20of%20%7DNa_3PO_4%7D%3D%5Cfrac%7B10.0g%7D%7B163.94g%2Fmole%7D%3D0.060moles)
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction will be,
![2MgCl_2(aq)+3Na_3PO_4(aq)\rightarrow 6NaCl(aq)+Mg_3(PO_4)_2(s)](https://tex.z-dn.net/?f=2MgCl_2%28aq%29%2B3Na_3PO_4%28aq%29%5Crightarrow%206NaCl%28aq%29%2BMg_3%28PO_4%29_2%28s%29)
From the balanced reaction we conclude that
As, 3 mole of
react with 2 mole of ![MgCl_2](https://tex.z-dn.net/?f=MgCl_2)
So, 0.060 moles of
react with
moles of ![MgCl_2](https://tex.z-dn.net/?f=MgCl_2)
From this we conclude that,
is an excess reagent because the given moles are greater than the required moles and
is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of ![Mg_3(PO_4)_2](https://tex.z-dn.net/?f=Mg_3%28PO_4%29_2)
From the reaction, we conclude that
As, 3 mole of
react to give 1 mole of ![Mg_3(PO_4)_2](https://tex.z-dn.net/?f=Mg_3%28PO_4%29_2)
So, 0.06 moles of
react to give
moles of ![Mg_3(PO_4)_2](https://tex.z-dn.net/?f=Mg_3%28PO_4%29_2)
Now we have to calculate the mass of ![Mg_3(PO_4)_2](https://tex.z-dn.net/?f=Mg_3%28PO_4%29_2)
![\text{ Mass of }Mg_3(PO_4)_2=\text{ Moles of }Mg_3(PO_4)_2\times \text{ Molar mass of }Mg_3(PO_4)_2](https://tex.z-dn.net/?f=%5Ctext%7B%20Mass%20of%20%7DMg_3%28PO_4%29_2%3D%5Ctext%7B%20Moles%20of%20%7DMg_3%28PO_4%29_2%5Ctimes%20%5Ctext%7B%20Molar%20mass%20of%20%7DMg_3%28PO_4%29_2)
![\text{ Mass of }Mg_3(PO_4)_2=(0.020moles)\times (262.87g/mole)=5.3g](https://tex.z-dn.net/?f=%5Ctext%7B%20Mass%20of%20%7DMg_3%28PO_4%29_2%3D%280.020moles%29%5Ctimes%20%28262.87g%2Fmole%29%3D5.3g)
Therefore, the mass of
produced are 5.3 grams.