Question

1a) Complete and balance the molecular equation, including phases, for the reaction of aqueous sodium sulfate, Na2SO4, and aqueous barium nitrate, Ba(NO3)2.

Na₂ SO₄ (aq) + Ba (NO₃)₂ (aq) --------> BaSO₄(s) + 2 NaNO₃ (aq)

*Enter the balanced net ionic equation, including phases, for this reaction.*

________________________________________ I Put - Ba²⁺ (aq) + SO₄^2- (aq) ------> BaSO₄ (s) but it is WRONG:(

1b)  The Ostwald process is used commercially to produce nitric acid, which is, in turn, used in many modern chemical processes. In the first step of the Ostwald process, ammonia is reacted with oxygen gas to produce nitric oxide and water. What is the maximum mass of H2O that can be produced by combining 62.6 g of each reactant? 4NH₃(g) + 5O₂(g) ---> 4NO(g) + 6H₂ o(g)

* Number*

* ______g H₂O*

  

Answer 1

Na₂ SO₄ (aq) + Ba (NO₃)₂ (aq) --------> BaSO₄(s) + 2 NaNO₃ (aq)

2Na^+(aq) + SO₄^2- (aq) + Ba^2+ (aq)+2 (NO₃)^- (aq) --------> BaSO₄(s) + 2 Na^+(aq) +2NO₃^- (aq)

removal of spectator ions to get net ionic equation

SO₄^2- (aq) + Ba^2+ (aq) --------> BaSO₄(s)   net ionic equation

1b.

    4NH₃(g) + 5O₂(g) ---> 4NO(g) + 6H₂O(g)

no of moles of NH3   = W/G.M.Wt

                                  = 62.6/17   = 3.68 moles

no of moles of O2    = 62.6/32     = 1.96 moles

4 moles of NH3 react with 5 moles of O2

3.68 moles of NH3 react with = 5*3.68/4   = 4.6 moles of O2 is required

O2 is limiting reactant

5 moles of O2 react with NH3 to gives 6 moles of H2O

1.96 moles of O2 react with NH3 to gives = 6*1.96/5    = 2.352 moles of H2O

mass of H2O   = no ofmoles * gram molar mass

                        = 2.352*18   = 42.336g of H2O >>>answer