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.

it is the most commonly accepted one