Question

2. 1.5 moles of AgNO3 reacts with 0.5 mole of Mg3P2. Calculate the moles of excess

Answer 1

Answer:

No of Moles in excess at the end of the reaction  is 0.25 moles

Explanation:

AgNO3  +   Mg3P2  → Ag3P + Mg(NO3)2

Balancing the equation we get

6AgNO3  +   Mg3P2  → 2Ag3P + 3Mg(NO3)2

6 moles of AgNO3 needs 1 mole of Mg3P2

using unitary method

AgNO3 = $\frac{1}{6}*Mg3P2$

1.5 AgNO3 = $\frac{1}{6}*1.5$

                  = 1/4 = 0.25moles of Mg3P2

So 1.5 Moles of AgNO3 requires 0.25Mg3P2 for complete reaction but we have 0.5Moles of Mg3P2 available Therefore Mg3P2 is in excess

No of Moles in excess at the end of the reaction = 0.5 - 0.25 = 0.25moles