Answer : The temperature in kelvins is, 
Explanation : Given,
= 178.5 KJ/mole = 178500 J/mole
= 161.0 J/mole.K
Gibbs–Helmholtz equation is :

As per question the reaction is spontaneous that means the value of
is negative or we can say that the value of
is less than zero.

The above expression will be:



Now put all the given values in this expression, we get :


Therefore, the temperature in kelvins is, 
Explanation:
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Answer:
8.8g of Al are necessaries
Explanation:
Based on the reaction, 2 moles of Al are required to produce 3 moles of hydrogen gas.
To solve this question we must find the moles of H2 in 11L at STP using PV = nRT. With these moles we can find the moles of Al required and its mass as follows:
<em>Moles H2:</em>
PV = nRT; PV/RT = n
<em>Where P is pressure = 1atm at STP; V is volume = 11L; R is gas constant = 0.082atmL/molK and T is absolute temperature = 273.15K at STP</em>
Replacing:
1atm*11L/0.082atmL/molK*273.15K = n
n = 0.491 moles of H2 must be produced
<em />
<em>Moles Al:</em>
0.491 moles of H2 * (2mol Al / 3mol H2) = 0.327moles of Al are required
<em />
<em>Mass Al -Molar mass: 26.98g/mol-:</em>
0.327moles of Al * (26.98g / mol) = 8.8g of Al are necessaries