Q1. Write balanced equationsfor the following reactions(a)Reaction between magne

Question

Q1. Write balanced equationsfor the following reactions(a)Reaction between magnesium(Mg) and nitrogen(N2) to form Mg3N2at high a temperature.

(b)Reaction between Mg3N2and waterto form magnesium hydroxideand ammonia(NH3)

(c)Decomposition of magnesium hydroxidewhen it is strongly heated.[See step 22 of the procedure for the products formed in this reaction.]

Q2.A sample of 0.756 molesof propane(C3H8) is completely burnedin 1.95 molesof oxygen. The products are carbon dioxideand water.

(a) Write thebalanced equation.

(b) Which is the limiting reactant? (Show calculations)

(c) How many moles of water are produced during this reaction?(Show calculations)

(d) Which is the excess reactant? How much of the excess reactant remains unreacted?

Explanation / Answer

Ans. #1.a. 3 Mg + N2 -----heat-----> Mg3N2

#b. Mg3N2 + 6 H2O -----------------> 3 Mg(OH)2 + 2 NH3

#c. Mg(OH)2 -----heat-----> MgO + H2O

#2.a. Balanced reaction:      C3H8 + 5 O2 ---------> 3 CO2 + 4 H2O

#b. Theoretical molar ratio of reactant:     C3H8 : O2 = 1 : 5

# Experimental molar ratio of reactant:     C3H8 : O2 = 0.756 mol : 1.95 mol = 1 : 2.58

Comparing the theoretical and experimental molar ratio of reactants, the experimental moles of O2 is less than its theoretical value of 5 mole, while that of propane is kept constant at 1 mol. Therefore, O2 is the limiting reactant.

# The formation of product follows the stoichiometry of limiting reactant.

Following stoichiometry of balanced reaction, 5 mol O2 produces 4 mol H2O.

So,

            Moles of H2O formed = (4/5) x moles of O2 = (4/5) x 2.58 mol = 2.064 mol

# Propane is the reagent in excess (see #b).

Following stoichiometry of balanced reaction, 5 mol O2 consumes 1 mol propane.

So,

            Moles of propane consumed formed = (1/5) x moles of O2

                                                                        = (1/5) x 2.58 mol

                                                                        = 0.516 mol

# Remaining moles of propane = Initial moles – Moles consumed

                                                            = 0.756 mol – 0.516 mol

                                                            = 0.240 mol

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