Question

the number of molecules of ammonia produced by 560 gram of dinitrogen gas​

Answer 1

Answer:

N2(g) + 3H2-> 2NH3(g)  This is the balanced equation

N2(g) + 3H2-> 2NH3(g)  This is the balanced equationNote the mole ratio between N2, H2 and NH3.  It is 1 : 3 : 2  This will be important.

N2(g) + 3H2-> 2NH3(g)  This is the balanced equationNote the mole ratio between N2, H2 and NH3.  It is 1 : 3 : 2  This will be important. 

N2(g) + 3H2-> 2NH3(g)  This is the balanced equationNote the mole ratio between N2, H2 and NH3.  It is 1 : 3 : 2  This will be important. moles N2 present = 28.0 g N2 x 1 mole N2/28 g = 1 mole N2 present

N2(g) + 3H2-> 2NH3(g)  This is the balanced equationNote the mole ratio between N2, H2 and NH3.  It is 1 : 3 : 2  This will be important. moles N2 present = 28.0 g N2 x 1 mole N2/28 g = 1 mole N2 presentmoles H2 present = 25.0 g H2 x 1 mole H2/2 g = 12.5 moles H2 present

N2(g) + 3H2-> 2NH3(g)  This is the balanced equationNote the mole ratio between N2, H2 and NH3.  It is 1 : 3 : 2  This will be important. moles N2 present = 28.0 g N2 x 1 mole N2/28 g = 1 mole N2 presentmoles H2 present = 25.0 g H2 x 1 mole H2/2 g = 12.5 moles H2 presentBased on mole ratio, N2 is limiting in this situation because there is more than enough H2 but not enough N2.

N2(g) + 3H2-> 2NH3(g)  This is the balanced equationNote the mole ratio between N2, H2 and NH3.  It is 1 : 3 : 2  This will be important. moles N2 present = 28.0 g N2 x 1 mole N2/28 g = 1 mole N2 presentmoles H2 present = 25.0 g H2 x 1 mole H2/2 g = 12.5 moles H2 presentBased on mole ratio, N2 is limiting in this situation because there is more than enough H2 but not enough N2. 

N2(g) + 3H2-> 2NH3(g)  This is the balanced equationNote the mole ratio between N2, H2 and NH3.  It is 1 : 3 : 2  This will be important. moles N2 present = 28.0 g N2 x 1 mole N2/28 g = 1 mole N2 presentmoles H2 present = 25.0 g H2 x 1 mole H2/2 g = 12.5 moles H2 presentBased on mole ratio, N2 is limiting in this situation because there is more than enough H2 but not enough N2. moles NH3 that can be produced = 1 mole N2 x 2 moles NH3/mole N2 = 2 moles NH3 can be produced 

N2(g) + 3H2-> 2NH3(g)  This is the balanced equationNote the mole ratio between N2, H2 and NH3.  It is 1 : 3 : 2  This will be important. moles N2 present = 28.0 g N2 x 1 mole N2/28 g = 1 mole N2 presentmoles H2 present = 25.0 g H2 x 1 mole H2/2 g = 12.5 moles H2 presentBased on mole ratio, N2 is limiting in this situation because there is more than enough H2 but not enough N2. moles NH3 that can be produced = 1 mole N2 x 2 moles NH3/mole N2 = 2 moles NH3 can be produced grams of NH3 that can be produced = 2 moles NH3 x 17 g/mole = 34 grams of NH3 can be produced

N2(g) + 3H2-> 2NH3(g)  This is the balanced equationNote the mole ratio between N2, H2 and NH3.  It is 1 : 3 : 2  This will be important. moles N2 present = 28.0 g N2 x 1 mole N2/28 g = 1 mole N2 presentmoles H2 present = 25.0 g H2 x 1 mole H2/2 g = 12.5 moles H2 presentBased on mole ratio, N2 is limiting in this situation because there is more than enough H2 but not enough N2. moles NH3 that can be produced = 1 mole N2 x 2 moles NH3/mole N2 = 2 moles NH3 can be produced grams of NH3 that can be produced = 2 moles NH3 x 17 g/mole = 34 grams of NH3 can be produced

Answer 2

Answer:

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