BaCl2+Al(NO3)3=Ba(NO2)3+AlCl3 How many moles of barium nitrate are produced if 4.25 moles of aluminum nitrate are used in the reaction?

6.75 moles of Ba(NO₃)₂.

Explanation:

The balanced equation for the reaction is given below:

3BaCl₂ + 2Al(NO₃)₃ —> 3Ba(NO₃)₂ + 2AlCl₃

From the balanced equation above,

2 moles of Al(NO₃)₃ reacted to produce 3 moles of Ba(NO₃)₂

Finally, we shall determine the number of mole of Ba(NO₃)₂ produced by the reaction of 4.25 moles of Al(NO₃)₃. This can be obtained as illustrated below:

From the balanced equation above,

2 moles of Al(NO₃)₃ reacted to produce 3 moles of Ba(NO₃)₂.

Therefore, 4.25 moles of Al(NO₃)₃ will react to produce = (4.50 × 3)/2 = 6.75 moles of Ba(NO₃)₂.

Thus, 6.75 moles of Ba(NO₃)₂ were obtained from the reaction.

the equation that correctly represent the reaction for formation of ammonia is

n2+ 3h2 → 2nh3 (answer d)

1 mole of nitrogen gas (n2) react with 3 moles of hydrogen gas (h2) to form 2 moles of ammonia ( nh3). this is in a process known as haber process were iron is used as a catalyst and reaction take place that a higher temperature and pressure. the process is exothermic hence energy is released.

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note 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

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based 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.