1.

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

Answer» <html><body><p> <br/> <br/> </p>Solution :A <a href="https://interviewquestions.tuteehub.com/tag/balanced-389334" style="font-weight:bold;" target="_blank" title="Click to know more about BALANCED">BALANCED</a> equation for the above reaction is written as follows, <br/> `N_(2(g)) + 3H_(2(g)) hArr 2NH_(3(g))`. <br/> Mole of `N_(2)=50.0 kg N_(2) xx (<a href="https://interviewquestions.tuteehub.com/tag/1000-265236" style="font-weight:bold;" target="_blank" title="Click to know more about 1000">1000</a> g N_(2))/(1 kg N_(2)) xx (1 "mol" N_(2))/(28.09g N_(2))` <br/> `= 17.86 xx 10^(2)` mol <br/> Mole of `H_(2) = 10.00 kg H_(2) xx (1000 g H_(2))/(1 kg H_(2)) xx (1 "mol" H_(2))/(2.016 g H_(2))` <br/> `= 4.96xx10^(3)` mol <br/> According to the equation ,1 mol `N_(2(g))` required 3 mol `(H_(2))`. For the reaction. <br/> Hence, for `17.86xx10^(2)` mol of `N_(2(g))` , the <a href="https://interviewquestions.tuteehub.com/tag/moles-1100459" style="font-weight:bold;" target="_blank" title="Click to know more about MOLES">MOLES</a> of `H_(2(g))` required would be : <br/> `= 17.86 xx 10^(2)` mol `xx(3 "mol"H_(2(g)))/(1 "mol"N_(2(g)))` <br/> `= 5.36xx10^(3)` mol `H_(2(g))` <br/> But we have only `4.96xx10^(3)` mol `H_(2)`. Hence diphydrogen is the limiting reagent in this <a href="https://interviewquestions.tuteehub.com/tag/case-910082" style="font-weight:bold;" target="_blank" title="Click to know more about CASE">CASE</a>. So, `NH_(3(g))` would be formed only from that amount of available dihydrogen , i.e `4.96xx10^(3)` mol `H_(2(g))`. <br/> Since 3 mol `H_(2(g)), 2 mol NH_(3(g))` <br/> `4.96xx10^(3) mol H_(2(g)) xx (2 mol NH_(3(g)))/(3 mol H_(2(g)))` <br/> `= 3.30xx10^(3) ` mol `NH_(3(g))` <br/> `3.30xx10^(3)` mol `NH_(3(g))` is obtained. <br/> If they are to be converted to <a href="https://interviewquestions.tuteehub.com/tag/grams-476111" style="font-weight:bold;" target="_blank" title="Click to know more about GRAMS">GRAMS</a>, it is done as follows: <br/> 1 mol `NH_(3(g))= 17.0 g NH_(3 (g))` <br/> `:.3.30xx10^(3) mol NH_(3(g))xx(17.0 g mol NH_(3(g)))/(1 mol NH_(3(g)))` <br/> `= 3.30xx10^(3) xx17 g NH_(3(g))` <br/> `= 56.1 xx 10^(3) g NH_(3)` <br/> `= 56.1 kg NH_(3(g))`</body></html>


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