Cao(s) + 2 nh4cl(s) = 2 nh3(g) + h2o(g) + cacl2(s) a) what is the mass of nh3 to be produced? (i think it's 68g but i'm not sure) b) how much the excess reactant remains after the reaction is complete?explain it to me, i need the answer asap though

This is stoichiometry question which involves unit conversions.

Given values:

CaO mass = 11.2g

NH4Cl mass = 22.4g

First convert the two given values to moles.

(11.2g CaO)(1mol CaO / 56.08g CaO) = 0.20mol CaO

(22.4g NH4Cl)(1mol NH4Cl / 53.49g NH4Cl) = 0.42mol NH4Cl

CaO reacts with NH4Cl on a 1:2 ratio.  This means that the reactants will react in this ratio: 0.20mol CaO: 0.40mol NH4Cl.  There will be an excess of 0.02mol NH4Cl unreacted because CaO is the limiting reagent (CaO is used up completely in the reaction and excess NH4Cl remains unreacted).

Now given the equation:

CaO(s) + 2 NH4Cl(s) = 2 NH3(g) + H2O(g) + CaCl2(s)

It is seen that every 2mol of NH4Cl reacts to form 2mol of NH3 because all the elements in the composition of NH3 is found in NH4Cl.  This means that NH4Cl and NH3 are on a 1:1 reaction ratio.  Now we use this relationship:

0.40mol NH4Cl reacted : 0.40mol NH3 produced.  Convert NH3 produced to grams to find solution:

(0.40mol NH3)(17.03g NH3 / 1mol NH3) = 6.8g NH3 produced a)

We determined the excess reactant as 0.02mol NH4Cl in a).  Now we convert it to grams:

(0.02mol NH4Cl)(53.49g NH4Cl / 1mol NH4Cl) = 1.07g NH4Cl unreacted b)