Answer:
21.5 g.
Explanation:
Hello!
In this case, since the reaction between the given compounds is:
[tex]2Li_3P+Al_2O_3\rightarrow 3Li_2O+2AlP[/tex]
We can see that according to the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction, the total mass of products equals the total mass of reactants based on the stoichiometric proportions; in such a way, we first need to compute the reacted moles of Li3P as shown below:
[tex]n_{Li_3P}^{reacted}=38gLi_3P*\frac{1molLi_3P}{51.8gLi_3P}=0.73molLi_3P[/tex]
Now, the moles of Li3P consumed by 15 g of Al2O3:
[tex]n_{Li_3P}^{consumed \ by \ Al_2O_3}=15gAl_2O_3*\frac{1molAl_2O_3}{101.96gAl_2O_3} *\frac{2molLi_3P}{1molAl_2O_3} =0.29molLi_3P[/tex]
Thus, we infer that just 0.29 moles of 0.73 react to form products; which means that the mass of formed products is:
[tex]m_{Li_2O}=0.29molLi_3P*\frac{3molLi_2O}{2molLi_3P} *\frac{29.88gLi_2O}{1molLi_2O} =13gLi_2O\\\\m_{AlP}=0.29molLi_3P*\frac{2molAlP}{2molLi_3P} *\frac{57.95gAlP}{1molAlP} =8.5gAlP[/tex]
Therefore, the total mass of products is:
[tex]m_{products}=13g+8.5g\\\\m_{products}=21.5g[/tex]
Which is not the same to the reactants (53 g) because there is an excess of Li₃P.
Best Regards!
