K2CO3(s) + 2HCl(aq) → 2KCl(aq) + H2O(l) + CO2(g)?
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Excess hydrochloric acid is reacted with a sample of potassium carbonate. After drying, the potassium chloride weighs 9.24 g. What is the volume of carbon dioxide gas produced if the reaction occurs at 15°C and 736 mmHg?
K2CO3(s) + 2HCl(aq) → 2KCl(aq) + H2O(l) + CO2(g)
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Calculate moles of KCl formed, and from that moles of CO2 that would have been produced:
9.24 g KCl X (1 mol / 74.56 g) X (1 mol CO2 / 2 mol KCl) = 0.0620 mol CO2 formed.
Then, use the ideal gas law to calculate the volume of that much CO2:
PV = n RT
P = 736/760 = 0.968 atm
T = 15 + 273 = 288 K
R = 0.0821 Latm/mol K
V = n RT/P
V = 0.0620(0.0821)(288)/0.968 = 1.51 L
From the balanced equation:
2mol KCl will be produced for every mol of CO2 produced
Molar mass KCl = 39.1+35.45 = 74.55g/mol
9.24g = 9.24/74.55 = 0.124 mol KCl
Therefore 0.124/2 = 0.062 mol CO2 will be produced
Determine volume using ideal gas equation:
PV = nRT
P = pressure = 736/760 = 0.968 atm
V = volume = what you want
n= moles = 0.062
R = gas constant = 0.082057
T = temp = 15+273 = 288K
0.968*V = 0.062 * 0.082057*288
V = 1.465/0.968
v = 1.51L CO2 produced.