Question

2A + N2 -> 2AIN How many moles of aluminum nitride form when 5.00 g of aluminum reacts with excess nitrogen? K 0.0925 mol 0 0.185 mol Oleo mol z se mol​

Answer 1

Answer:

0.185 moles of AlN

Explanation:

The equation is:

2Al  +  N₂  →  2AlN

This reaction indicates that 2 mol of aluminum can react to 1 mol of nitrogen to produce 2 moles of aluminum nitride.

We convert the aluminum's mass to moles

5 g/ 26.98 g/mol = 0.185 moles.

2 moles of Al are needed to produce 2 moles of AlN

so, ratio is 1:1

In conclussion we say:

0.185 moles of Al, must produce 0.185 moles of AlN