What is the mass of the NH4Cl produced

What is the mass of the solid NH4Cl formed when 68.5 g of NH3 is mixed with an equal mass of HCl?

What is the volume of the gas remaining, measured at 287.15 kelvin and 0.989 atm?

which gas remains? (NH3 or HCl)

NH3 (gas) +HCl (gas) --> NH4Cl (solid)

please help me. thanks

NH3(g) + HCl(g) --> NH4Cl(s)
68.5g .... 68.5g ....... ???g

This is a limiting reactant question, and one way to solve it is to compute the mass of the product using each reactant. The smaller amount of product is the actual amount, and the reactant giving the smaller amount is the limiting reactant.

68.5 NH3 x (1 mol NH3 / 17.0g NH3) x (1 mol NH4Cl / 1 mol NH3) x (53.5g NH4Cl / 1 mol NH4Cl) = 215.6 g NH4Cl

68.5 HCl x (1 mol HCl / 36.5g HCl) x (1 mol NH4Cl / 1 mol NH3) x (53.5g NH4Cl / 1 mol NH4Cl) = 100.4 g NH4Cl

100.4 g of NH4Cl is produced. HCl is the limiting reactant, and NH3 is in excess. The remaining NH3 will provide some gas pressure in the vessel.

Moles of NH3 remaining:
68.5 g NH3 x (1 mol NH3 x 17.0g NH3) = 4.03 mol HCl
68.5g HCl x (1 mol HCl / 36.5 g HCl) = 1.88 mol HCl
1.88 mol NH3 will react, leaving 2.15 mol NH3 unreacted

PV = nRT
V = nRT / P
V = 2.15 mol x 0.0821 Latm/molK x 287.15K / 0.989 atm
V = 51.3 L