If 33.7 g of Ag_2 O(s) are mixed with 46.5 g of C_10 H_10 N_4 SO_2 (s) and the r

Question

If 33.7 g of Ag_2 O(s) are mixed with 46.5 g of C_10 H_10 N_4 SO_2 (s) and the reaction is run. answer the following questions. (a) What is the limiting reagent? (b) Assuming no side reactions, how much H_2 O (I) can be produced under these conditions? (c) If 1 21 g of H_2 O (g) are produced in the lab under these conditions, what is the percent yield? (d) How many grams of the reactant in excess remains at the end of the experiment? Data Atomic mass Ag = 107.9, N = 14.01, H = 1.008, O = 16.00, S = 32.07, C 12.01

Explanation / Answer

2 C10 H10 N4 SO2 + Ag2O 2 Ag C10 H9 N4 SO2+H2O

a)

MW of Ag2O = 2 * 107.9 + 16 = 231.8 g/mol

Moles of Ag2O = 33.7 / (MW of Ag2O) = 33.7 / 231.8

= 0.145

MW of C10 H10 N4 SO2 = 10 * 12.01 + 10 * 1.008 + 4 * 14.01 + 32.07 + 2 * 16 = 250.29 g/mol

Moles of C10 H10 N4 SO2 = 46.5 / (MW of Ag2O) = 46.5 / 250.29

= 0.186

1 mole of Ag2O reacts with 2 moles of C10 H10 N4SO2

2 * Moles of Ag2O > Moles of C10 H10 N4 SO2

So, by stoichiometry C10 H10 N4 SO2 is the limiting reagent.

b)

MW of H2O = 2 * 1.008 + 16 = 18.016 g/mol

Moles of H2O produced = Moles of C10 H10 N4 SO2reacted / 2

Moles of H2O produced = 0.186 /2 = 0.093

Mass of H2O produced = 0.093 * 18.016 = 1.67 g

c)

% Yield = Actual mass of H2O produced / Calculated mass of H2O * 100

= 1.21 / 1.67 * 100 = 72.3 %

d)

Moles of H2O produced = 1.21 / 18.016 = 0.067

Moles of C10 H10 N4 SO2 reacted = 2 * Moles of H2O produced

= 0.134

Moles of excess C10 H10 N4 SO2 remaining = 0.186 - 0.134 = 0.051

Mass of excess C10 H10 N4 SO2 remaining = 0.051 * 250.29 g

= 12.88 g

Moles of Ag2O reacted = Moles of C10 H10 N4 SO2reacted / 2

= 0.134 / 2 = 0.067

Moles of excess Ag2O remaining = 0.145 - 0.067 = 0.078

Mass of excess Ag2O remaining = 0.078 * 231.8 g

= 18.13 g

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