Calculate the % MgCO, in the mixture. 2) A student received a mixture containing

Question

Calculate the % MgCO, in the mixture. 2) A student received a mixture containing KCI/KCIO3 and Mn02. She analyzed the sample and obtained the following data. the following data: Weight of test tube and contents before heating Weight of test tube after heating Weight of test tube Volume of water displaced Barometric reading Oxygen temperature Vapor pressure of water 61.88 g 60.70 g 34.38 g 1445.0 mls 75.21 cm Hg 25.3°C 31 torr 2 KCIO, 2 KCI + 302 Calculate the percent KCIO, in the mixture using the above volume of oxygen data 3) Calculate the total heat change when 100.0 g of ice water at -20°C was heated to steam water at 120.0oC.

Explanation / Answer

2.

R = 0.0821 L.atm.K-1.mol-1

Temperature of oxygen = 25.3 + 273.15 = 298.45 K

Volume of oxygen = 1445.0 mL = 1.445 L

Pressure of oxygen = 752.1 - 31 = 721.1 torr = 721.1 / 760 atm = 0.949 atm

Using ideal gas equation,

P V = n R T
n = 0.949 * 1.445 / (0.0821 * 298.45)

n = 0.0560 mol of oxygen liberated.

From the balanced equation,

3 mol of oxygen is liberated from 2 mol of KClO3

Then, 0.0560 mol of O2 is liberated from = 2 * 0.0560 / 3 = 0.0373 mo of KClO3

Mass of KClO3 = moles * molar mass = 0.0373 * 122.5 = 4.57 g.

mass of mixture of before heating = 61.88 - 34.38 = 27.5 g.

Therefore,

Mass % of KClO3 = mass of KClO3 * 100 / mass of mixture = 4.57 * 100 / 27.5 = 16.6 %

3.

Heat change = mass * specific heat * change in temperature

q1 = 100 * 2.108 * (0 - (- 20)) = 4216 J

q2 = heat of fusion of 100 g. of ice = 100 * 334 = 33400 J

q3 = 100 * 4.184 * (100 - 0) = 41840 J

q4 = heat of vapourisation of 100 g. of water = 100 * 2257 = 225700 J

q5 = 100 * 1.996 * (120.0 - 100) = 3992 J

Therefore,

Total heat change = 4216 + 33400 + 41840 + 225700 + 3992 = 309148 J = 309.148 kJ

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