Question

a mixture of 73.0 g NH3 and 73.0 g HCl are made to react at stp according to the equation NH3 + HCl --> NH4Cl What is the volume of gas remaining and what gas is it?

Answer 1

m = 73 g of NH3

MW NH3 = 17

mol = 73/17 = 4.294117 mol o fNH3

m = 73 g of HCl

MW HCl = 36

mol of HCl = 73/36 = 2.0277 mol of HCl

2 mol of gas form --> 1 mol of gas

2.0277 molof HCl react with 2.0277 mol of NH3 to form 2.0277 mol of NH3

then

mol of HCl left = 2.0277 - 2.0277 = 0

mol of NH3 left = 4.294117 - 2.0277 =2.266417

mol of NH4cl formed = 2.0277

total final mol = 2.0277 +2.266417 = 4.294117

so reduction of volume:

Vf = 4.294117/(4.294117 +2.0277 ) = 0.67925Vi

If we assume this was at STP

then

1 mol = 22.4 L

that is

initial mol = 2.0277 +4.294117 =6.321817

initial V = 22.4*6.321817 = 141.6087008 L

then

since

Vf = 0.67925Vi

Vf = 0.67925 *141.6087008 = 96.187710 L