3Si + 2N2 --> Si3N4 (as given)
n(Si) = m/MM = 38.25/28.085 = 1.3619 mol
n(N2) = 14.33/2*14.007 = 0.5115 mol
Therefore, N2 is limiting and Si is in excess
The molar ratio of 2N2:Si3N4 is 2:1
So, 0.0575 mol of silicon nitride is formed (dividing 0.5115 by 2)
m of silicon nitride= n*mm = 0.0575*140.283 = 8.06627... g
= 8.066g (4 significant figures)
(hopefully it is right, but double check in case i did something wrong) :)
Molarity = Moles of solute/ L(liters) of solution
So let's plug in the information.
5.0 moles/10L = 0.5 M
Q = ?
Cp = 0.450 j/g°C
Δt = 49.0ºC - 25ºC => 24ºC
m = 55.8 g
Q = m x Cp x Δt
Q = 55.8 x 0.450 x 24
Q = 602.64 J
hope this helps!
