Respuesta :
Answer: 6 moles of KNO3 could be produced when 6 moles of K3PO4 and 2 moles of Al(NO3)3 react
Explanation:
The question requires us to determine the amount of potassium nitrate (KNO3), in moles, produced in the following reaction when 6 moles of potassium phosphate (K3PO4) react with 2 moles of aluminum nitrate (Al(NO3)3):
[tex]K_3PO_4+Al\left(NO_3\right)_3\rightarrow3KNO_3+AlPO_3[/tex]Since the amount of both reactants was provided by the question, we need to determine the limiting reactant before calculating the amount of product obtained.
According to the balanced chemical equation provided, 1 mol of K3PO4 is necessary to react with 1 mol of Al(NO3)3. Thus, we can determine how many moles of Al(NO3)3 would be necessary to react with 6 moles of K3PO4:
1 mol K3PO4 ----------------- 1 mol Al(NO3)3
6 mol K3PO4 --------------- x
Solving for x, we'll have:
[tex]x=\frac{(1\text{ mol Al\lparen NO}_3)_3)\times(6\text{ mol K}_3PO_4)}{(1\text{ mol K}_3PO_4)}=6\text{ mol Al\lparen NO}_3)_3[/tex]Therefore, 6 moles of Al(NO3)3 would be necessary to react completely with the given amount of K3PO4. Since the given amount of Al(NO3)3 (2 mol) is less than the required amount (6 mol), we can say that Al(NO3)3 is the limiting reactant.
Next, we can determine the amount of KNO3 formed from the given amount of Al(NO3)3, our limiting reactant.
From the balanced chemical equation, we can see that 1 mol of Al(NO3)3 is necessary to produce 3 moles of KNO3. Thus, we can write:
1 mol Al(NO3)3 --------------------- 3 mol KNO3
2 mol Al(NO3)3 -------------------- y
Solving for y, we'll have:
[tex]y=\frac{(2\text{ mol Al\lparen NO}_3)_3)\times(3\text{ mol KNO}_3)}{(1\text{ mol Al\lparen NO}_3)_3)}=6\text{ mol KNO}_3[/tex]Therefore, 6 moles of KNO3 could be produced when 6 moles of K3PO4 and 2 moles of Al(NO3)3 react.
