Question

1. A 4.385 g sample of mercury (II)oxide was decomposed to produce 4.003 g of liquid mercury according to the equation below. Calculate the percent yield of mercury. 2 HgO(s)-> 2Hg() + O2(g) 4.385 (21656 =4.0039 2. Strontium carbonate decomposes into strontium oxide and carbon dioxide gas when heated, as shown below. What mass of strontium carbonate would be required to produce 49 g of strontium oxide, assuming the reaction went at 100% yield? SrCO3(s) > SEO (s) + CO2(g) 3. What volume of oxygen, in ml, would be produced at STP (0 °C and 1.00 atm) by the decomposition of 3.428 g of sodium chlorate, according to the equation below? 2 NaClO,(s) -> 2 NaCl(s) + 302() 4. If a 2.465 g sample of potassium hydrogen carbonate lost 0.798 g of mass during a decomposition reaction, what was the % yield of the reaction? 2 KHCO3(s) ->K,CO3(s) + H2O(g) + CO2(g)

Answer 1

1. Moles of HgO = mass of HgO / molar mass of HgO = 4.385 / 216.6 = 0.020 moles

Moles of Hg theoretically formed = moles of HgO reacted = 0.02 moles

Mass of Hg theoretically formed = moles * molar mass = 0.02 * 200.6 = 4.012 g

% yield =practical mass formed / theoretical mass * 100 = 4.003 / 4.012 * 100 = 99.78%