Answer : The volume of reactant measured at STP left over is 409.9 L
Explanation :
First we have to calculate the moles of
and
by using ideal gas equation.
<u>For
:</u>

where,
P = Pressure of gas at STP = 1 atm
V = Volume of
gas = 1580 L
n = number of moles
= ?
R = Gas constant = 
T = Temperature of gas at STP = 273 K
Putting values in above equation, we get:


<u>For
:</u>

where,
P = Pressure of gas at STP = 1 atm
V = Volume of
gas = 3510 L
n = number of moles
= ?
R = Gas constant = 
T = Temperature of gas at STP = 273 K
Putting values in above equation, we get:


Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,

From the balanced reaction we conclude that
As, 3 mole of
react with 1 mole of 
So, 156.6 moles of
react with
moles of 
From this we conclude that,
is an excess reagent because the given moles are greater than the required moles and
is a limiting reagent and it limits the formation of product.
Now we have to calculate the excess moles of
reactant (unreacted gas).
Excess moles of
reactant = 70.49 - 52.2 = 18.29 moles
Now we have to calculate the volume of reactant, measured at STP, is left over.

where,
P = Pressure of gas at STP = 1 atm
V = Volume of gas = ?
n = number of moles of unreacted gas = 18.29 moles
R = Gas constant = 
T = Temperature of gas at STP = 273 K
Putting values in above equation, we get:


Therefore, the volume of reactant measured at STP left over is 409.9 L