Aluminum nitrite (al(no2)3) and ammonium chloride (nh4cl) react according to the following equation. al(no2)3(aq) + 3 nh4cl (aq) alcl3(aq) + 3 n2(g) + 6 h2o (l)how many grams of the excess reagent are left over when 62.5 g of aluminum nitrite reacts with 54.6 g of ammonium chloride? ?

m = 7.596 g

Explanation:

First thing we need to do is write the equation and balance if it's neccesary:

Al(NO2)3(aq) + 3NH4Cl(aq) > AlCl3(aq) + 3N2(g) + 6H2O(l)

We can see that this reaction has no need to be balanced.

Now, let's see who is the limiting reagent and the excess reagent. To do so, we need to calculate the moles of each reactant, using the following expression:

n = m/MM

MM is molecular mass and m is mass

The molecular mass is usually given, but if not, we can look for them.

In the case of Al(NO2)3 is 212.996 g/mol;

In the case of Ammonium chloride is 53.49 g/mol

Let's calculate the moles:

n1 = 62.5 / 212.996 = 0.293 moles

n2 = 54.6 / 53.49 = 1.021 moles

Now that we have the moles, let's get the ratio of mole between these two reactants. We can see this by looking the general reaction and it's coefficients (which represents theorical moles of each agent).

moles Al(NO2)3 / moles NH4Cl = 1/3

Now replacing one of the obtained moles above, we can see how much do we actually need to this reaction be taking place.

1 mole Al(NO2)3 > 3 moles NH4Cl

0.293 moles > x

Solving for x, will give the moles of NH4Cl needed to react with 0.293 moles of Al(NO2)3 so:

x = 0.293 * 3 / 1 = 0.879 moles NH4Cl

As we can see we only need 0.879 moles of NH4Cl, and we have 1.021 moles, therefore, the excess reagent is the NH4Cl

Now, the remanent moles of NH4Cl will be:

moles NH4Cl = 1.021 - 0.879 = 0.142 moles

Finally, to get the grams:

m = n * MM

solving for m:

m = 0.142 * 53.49

m = 7.596 g

the answer is 75.0%.