Calculate the number of grams of excess reactant

Ammonia rapidly reacts with hydrogen chloride, making ammonium chloride. Calculate the number of grams of excess reactant when 7.40 g of NH3 reacts with 4.00 g of HCl.

I really need help with this problem. I have no idea how to calculate this. When I attempted to solve the problem, the first thing I did was write the balanced chemical equation:

NH3 + HCl ----> NH4Cl

Then, I converted 7.40 grams of NH3 and 4.00 grams of HCl into moles:

7.40 g (1 mol/17.03 g) = 0.4345 moles NH3
4.00 g (1 mole/36.46 g) = 0.1097 moles HCl

Based on this information, I can conclude that HCl is the limiting reactant, right? The problem is, I don't know where to go from here! Please help! It would be greatly appreciated!

NH3 + HCl --> NH4Cl
7.4gNH3 / 17g/mole = 0.435moles NH3
4g HCl / 36.45g/mole = 0.11moles HCl
reaction is 1:1
HCl = limiting reactant using only 0.11moles NH3
0.435molesNH3 - 0.11moles NH3 used = 0.325moles NH3 left over
0.325moles NH3 x 17g/mole = 5.52g