Ammonia gas can be prepared by the reaction according to the following equation:

Question

Ammonia gas can be prepared by the reaction according to the following equation: [Mwt of (CaO = 56.08 g/mol; NH4Cl = 53.5 g/mol; NH3 = 17.03 g/mol; H2O = 18.02 g/mol; CaCl2 = 110.98 g/mol;). How many moles of water can be formed If 112 g CaO reacts with 224 g NH4CI? How many grams of Cacl2 can be formed If112 g CaO reacts with 224 g NH4Cl? How many grams of NH3 gas can be formed from If 112 g CaO reacts with 224 g NH4Cl? If this reaction produces 31 g of NH3 gas when 112 g CaO reacts with 224 g NH4Cl. How much the percents yield? How many grams of the limiting reactant required reacting with the excess material?

Explanation / Answer

Moles of CaO= 112/56= 2

Moles of NH4Cl= 224/53.5=4.18

1)As 2 moles of CaO requires only 4 moles of CaO, thus 2 moles of CaO will react. As one mole of CaO produces one mole of H2O

2 moles produce 2 moles of h2O

2) 2 moles of CaCl2 are produced

mass= 2*110.98 =222 gm

3)moles of NH3 produced=4

==> mass=4*17.03= 68 gm

4) As we saw in question 3, the yeild was supposed to be 68 gm, but it is only 31 gm

thus % yeild= (31/68)*100= 45.58%

5) As 4 moles of NH4Cl were used,==> 4*53.5= 214 gms were used

==> excess= 224-214=10 gms

==>moles of NH4Cl=10/53.5

==> moles of Cao required= (10/53.5)/ 2.....asa every 2 moles of NH4Cl requires 1 mole of CaO

==> mass=(5/53.5)*56 = 5.233 gms

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