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,' owLv2} Unine tes X Ceramic penodic Table i a s For the following reacti e Chegg Study! Guided. + w.cornm/takceAssignmenttakeCovalentActivity.do locatorsassignment-take&takeAssignmentSessionl; req Use the References to access important valaes if eeded for thias qaestion For the following reaction, 447 grams of nitrogen gas are allowed to react with 5.67 grams of hydrogen gas nitrogen (g) + hydrogen (g)--ammonia (g) What is the maximam amount of ammonia that can be formed? grams What is the FORMULA for the limiting reagent crams What amount of the excess reagent remains after the reaction is complete? Retry Entire Grop more group attempts remaining Subrmit Answer The equation for this reaction is: Na (g) + 3H2 (g)- 2NH3 (g) D E DOLLExplanation / Answer
N2(g) + 3H2(g) ------------------> 2NH3(g)
no of moles of N2 = W/G.M.Wt
= 4.47/28 = 0.16 moles
no of moles of H2 = W/G.M.Wt
= 5.67/2 = 2.835 moles
N2(g) + 3H2(g) ------------------> 2NH3(g)
3 moles of H2 react with 1 moles of N2
2.835 moles of H2 react with = 1*2.835/3 = 0.945 moles of N2
N2 is limiting reactant
1 moles of N2 react with H2 to gives 2 moles of NH3
0.16 moles of N2 react with H2 to gives = 2*0.16/1 = 0.32 moles of NH3
mass of NH3 = no of moles * gram molar mass
= 0.32*17 = 5.44g of NH3
The maximum amount of NH3 = 5.44g
1 moles of N2 react with 3 moles of H2
0.16 moles of N2 react with = 3*0.16/1 = 0.48 moles of H2 is required
H2 is excess reactant
The no of moles of excess reactant remains after complete the reaction = 2.835-0.48 = 2.355 moles
The amount of excess reactant remains after complete the reaction = no of moles *gram molar mass
= 2.355*2 = 4.71g
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