Date: Section: Name: Post-lab Questions Q1. Write balanced equations for the fol

Question

Date: Section: Name: Post-lab Questions Q1. Write balanced equations for the following reactions (a) Reaction between magnesium (Mg) and nitrogen (N) to form Mg N at high a temperature. (b) Reaction between Mg>N and water to form magnesium hydroxide and ammonia (NH,) (e) Decomposition of magnesium hydroside when it is strongly heated. [See step 22 of the procedure for the products formed in this reaction.] 02. A sample of 0.756 moles of propane (C,Hs) is completely burned in 1.95 moles of oxygen. The products are carbon dioxide and water. (a) Write the balanced 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

1. (a) 3Mg +N2 ----> Mg3N2

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

   (c) Mg(OH)2     -------> MgO + H2O

2. (a) C3H8 + 5O2 -----> 3CO2 + 4H2O

    (b) From above reaction for complete combution of 1 mole of C3H8 5 moles of O2 is required. Therefore, for 0.756 moles of C3H8 one would need 0.756*5= 3.78 moles of O2.

But given, 1.95 moles of O2 can cause the combustion of 1.95/5= 0.39 moles of C3H8.

Hence, limiting reactant is oxygen.

(c) From the balance chemical reaction for the combustion of C3H8. 5 moles of O2 will produce 4 moles of water.

Hence, 1.95 moles of O2 will produce (4/5)*1.95 = 1.56 moles of water

(d) Excess reactant is C3H8.

Since, 1.95 moles of O2 can consume 0.39 moles of C3H8 (as calculated in 2(b)).

Initial amount of C3H8 = 0.756 moles

Hence, unreacted C3H8 = 0.756 - 0.39 = 0.366 moles.

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