1.Calculate the mass of CaCO3 that would produce ~ 40.0 ml of CO2 at STP.
Hint use molar volume of an ideal gas (22.4 L).
2. The pressure of water vapor at 23 °C is [_____] torr.
3.The difference in mass of the gas generator before and after the reaction takes place yields the mass of CO2 evolved.
true
false
Q1
CaCO3 + 2HCl = H2O + CO2 + CaCL2
40 mL of CO2 at STP --> 40*10^-4 L
1 mol = 22.4 L at STP
x mol = 40*10^-4 L
x = (40*10^-3)/(22.4) = 0.0017857 mol of CO2
ratio is 1:1 so
0.0017857 mol of CO2 =0.0017857mol of CaCO3
mass = mol*MW= 0.0017857*100 = 0.17857 g of CaCO3 wil do
Q2
23°C the water will have a vapor pressure of --> P = 21.1 torr (from data bases)
Q3
Correct
since only CO2 is being "lost" i.e. the solid/liquyd solutoin remains with CaCO3, CaCl2 and H2O
only CO2(g) goes "out"
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