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28 g of nitrogen and 6 g hydrogen were mixed in a1 litre closed container. At equilibrim 17 g NH_3 was produced. Calculate the weight of weight of nitrogen, hydrogen at equilibrium.

Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/given-473447" style="font-weight:bold;" target="_blank" title="Click to know more about GIVEN">GIVEN</a> `m_(N_2) = <a href="https://interviewquestions.tuteehub.com/tag/28g-1834386" style="font-weight:bold;" target="_blank" title="Click to know more about 28G">28G</a>, m_(H_2) = 6 g , V = 1L`<br/> `(n_(N_2))_("initial")= <a href="https://interviewquestions.tuteehub.com/tag/28-299271" style="font-weight:bold;" target="_blank" title="Click to know more about 28">28</a>/28 = 1mol` <br/> `(n_(H_2))_("initial")= 6/2 = 3mol`<br/> `N_2(g)+ 3H_2(g) hArr 2NH_3(g)`<br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/FM_CHE_XI_V02_C08_E01_046_S01.png" width="80%"/> <br/> `[NH_3] = (17/17) = 1mol = 2x`<br/> `rArrx = 0.5` mol <br/> At equilibrium, `[N_2] = 1 - x = 0.5 mol`<br/> `[H_2] = 3 -3x =3-3 (0.5) = 1.5 mol` <br/> <a href="https://interviewquestions.tuteehub.com/tag/weight-1451304" style="font-weight:bold;" target="_blank" title="Click to know more about WEIGHT">WEIGHT</a> of`N_2` = (no. of moles of `N_2`)` xx`molar <a href="https://interviewquestions.tuteehub.com/tag/mass-1088425" style="font-weight:bold;" target="_blank" title="Click to know more about MASS">MASS</a> of `N_2 = 0.5 xx 28 = 14g` <br/> Weight of `H_2` = (no. of molesof `H_2`) `xx` molar mass of `H_2= 1.5 xx 2 = 3g`</body></html>


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