In order to produce Hg2Br2, the following procedure is carried out: A 30.0 mL sa
ID: 748302 • Letter: I
Question
In order to produce Hg2Br2, the following procedure is carried out: A 30.0 mL sample of aqueous 0.300 M Hg2(NO3)2 is mixed with 25.0 mL of an aqueous 0.500 M NaBr solution. What is the maximum mass of Hg2Br2 that could be produced from this reaction? Balanced Reaction: Hg2(NO3)2 (aq) + 2NaBr (aq) ---------------> Hg2Br2 (s) + 2NaNO3 (aq)Explanation / Answer
Balance Equation: Mg3N2(s) + 6H2O(l) --> 3Mg(OH)2(s) + 2NH3(aq) Find out how many moles of substance: Mg3N2 = 30.5g/100.95 g/mol = 0.302mol H2O = 28.0g/18.015 g/mol = 1.554mol Limiting reagent is H2O (because you need 6x as many moles as Mg3N2) Every mole of Mg3N2 produces 2 Moles of NH3. Thus, 0.302mol x 2 = 0.604 mols NH3 Molar mass of NH3 = 17.031 g/mol 0.604 x 17.031 g/mol = 10.29g NH3