How many liters of ammonia are formed with 49.8 g of nitrogen gas reacting with 10.7 g of hydrogen gas at S. T.P.?

The reaction is as follows:

N2+3H2>2NH3

the number of moles of N2 is 49.8g/14gmol^-1=3.56mol

the number of moles of H2 is 10.7g/2gmol^-1=5.35mol

here, since the ratio of N2/H2 is 1/3, we can see that H2 is the limiting reagent. therefore, the reaction proceeds as follows:

N2 + 3H2 > 2NH3
5.36mol (2/3)(5.36mol)=3.57mol

therefore, the number of moles of ammonia formed is 3.57mol. at STP, P=1atm, T=273K.
Therefore, V(NH3)=(3.57)(22.38)=79.822 L

The answer is the first option just took the test