28 g of nitrogen and 6 g of hydrogen were mixed in a 1 litre closed container. At equilibrium 17 g NH_(3) was produced. Calculate the weight of nitrogen, hydrogen at equilibrium.

28 g of nitrogen and 6 g of hydrogen were mixed in a 1 litre closed co

1.	28 g of nitrogen and 6 g of hydrogen were mixed in a 1 litre closed container. At equilibrium 17 g NH_(3) was produced. Calculate the weight of nitrogen, hydrogen at equilibrium.
Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :<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))=6g,` <br/> `V=1L` <br/> `(n_(N_(2)))_("<a href="https://interviewquestions.tuteehub.com/tag/initial-499269" style="font-weight:bold;" target="_blank" title="Click to know more about INITIAL">INITIAL</a>")=(28)/(28)="1 mol"` <br/> `(n_(H_(2)))_("Initial")=6/2=3" mol"` <br/> `N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g)` <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/SUR_CHE_XI_V02_C08_E01_048_S01.png" width="80%"/> <br/> `[NH_(3)]=((17)/(17))="1 mol"` <br/> Weight of `N_(2)` = (no. of moles of `N_(2)` ) `xx` molar mass of `N_(2)` <br/> `=0.5xx28=14g` <br/> Weight of `H_(2)` = (no. of moles of `H_(2)` ) `xx` molar mass of `H_(2)` <br/> `=1.5xx2=3g`</body></html>