1.

50.0 kg of N_(2)(g) and 10.0 kgof H_(2)(g) are mixed to produce NH_(3)(g). Calculate mass of NH_(3)(g) formed. Identify the limiting reagent in the production of NH_(3) in this situation.

Answer» <html><body><p></p>Solution :The balanced chemical equation for the reaction is : <br/> `underset("1 <a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a>")(N_(2)(<a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a>))+underset("3 mol")(3H_(2)(g))rarrunderset("2 mol")(2NH_(3)(g))` <br/> <a href="https://interviewquestions.tuteehub.com/tag/step-25533" style="font-weight:bold;" target="_blank" title="Click to know more about STEP">STEP</a> I. To find the limiting reactant <br/> Mass of `N_(2)` taken `= 50.0 <a href="https://interviewquestions.tuteehub.com/tag/kg-1063886" style="font-weight:bold;" target="_blank" title="Click to know more about KG">KG</a> = 50,000 g` <br/> Moles of `N_(2)=("Mass of" N_(2))/("<a href="https://interviewquestions.tuteehub.com/tag/molar-562965" style="font-weight:bold;" target="_blank" title="Click to know more about MOLAR">MOLAR</a> mass of "N_(2))=(5000g)/((28"g mol"^(-1)))=1785.7mol` <br/> Mass of `H_(2)` taken `= 10.0 kg = 10,000 g` <br/> Moles of `H_(2)=(10,000g)/((2.0"g mol"^(-1)))=5000 mol` <br/> From the above balanced equation, 1 mol of `N_(2)` reacts with 3 mol of `H_(2)` <br/> `:. 1785.7` mol of `N_(2)` will react with` H_(2)=3xx1785.7 mol = 5355.9 mol` But we have only 5000 mol of `H_(2)`. Therefore, `H_(2)` is the limiting reactant. <br/> Step II. To find the amount of `NH_(3)` formed. <br/> 3 moles of `H_(2)` from `NH_(3) =2` mol <br/> 5000 moles of `H_(2)` form `NH_(3)=5000 xx ((2 mol))/((3 mol))=3333.3 mol` <br/> Mass of `NH_(3)` formed `= 3333.3 xx 17 = 56666.6 g = 56.67 kg`.</body></html>


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