assume 340 gram of ammonia need to be produced
1.how many moles of hydrogen are needed
2.how many grams of nitrogen needed.
Answers
Explanation:
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
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 NOTE: The key to this problem is recognizing that N2 is limiting, and therefore limits how much NH3 can be produced.