Question

I NEED HELP PLEASE!!!! CHEMISTRY QUESTION: If 38 g of Li3P and 15 grams of Al2O3 are reacted, what total mass of products will result? Why?

Answer 1

21.5 g.

Step-by-step explanation:

Hello!

In this case, since the reaction between the given compounds is:

`2Li_3P+Al_2O_3\rightarrow 3Li_2O+2AlP`

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:

`n_(Li_3P)^(reacted)=38gLi_3P*(1molLi_3P)/(51.8gLi_3P)=0.73molLi_3P`

Now, the moles of Li3P consumed by 15 g of Al2O3:

`n_(Li_3P)^(consumed \ by \ Al_2O_3)=15gAl_2O_3*(1molAl_2O_3)/(101.96gAl_2O_3) *(2molLi_3P)/(1molAl_2O_3) =0.29molLi_3P`

Thus, we infer that just 0.29 moles of 0.73 react to form products; which means that the mass of formed products is:

`m_(Li_2O)=0.29molLi_3P*(3molLi_2O)/(2molLi_3P) *(29.88gLi_2O)/(1molLi_2O) =13gLi_2O\\\\m_(AlP)=0.29molLi_3P*(2molAlP)/(2molLi_3P) *(57.95gAlP)/(1molAlP) =8.5gAlP`

Therefore, the total mass of products is:

`m_(products)=13g+8.5g\\\\m_(products)=21.5g`

Which is not the same to the reactants (53 g) because there is an excess of Li₃P.

Best Regards!