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 = 
1.5 AgNO3 = 
= 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
Answer:
(a). 132 × 10^-9 s = 132 nanoseconds.
(b)..176.5 pico-seconds.
Explanation:
(a). At one torr, the first thing to do is to find the speed and that can be done by using the formula below;
Speed = [ (8 × R × T)/ Mm × π]^1/2.
Where Mm = molar mass, T = temperature and R = gas constant.
Speed= [ ( 8 × 8.314 × 300)/ 131.293 × π × 10^-3)^1/2. = 220m/s.
The next thing to do now is to calculate for the degree of collision which can be calculated by using the formula below;
Degree of collision = √2 × π × speed × d^2 × pressure/ K × T.
Note that pressure = 1 torr = 133.32 N/m^2 and d = collision diameter.
Degree of collision = √2 × π × 220 × (4.9 × 10^-10)^2 × 133.32/ 1.38 × 10^-23 × 300.
Degree of collision = 7.55 × 10^6 s^-1.
Thus, 1/ 7.55 × 10^6. = 132 × 10^-9 s = 132 nanoseconds.
(b). At one bar;
1/10^5 × 10^3 × 56.65 = 1.765 × 10^-10 = 176.5 pico-seconds.
