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mmmeraki
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Hi! I got a question. I have been given the reaction of the equation and no other data. I have to calculate the mass percentage of H2CO3. The gas dissolves in water.
1. Homework Statement
CaCO3 + H2SO4 = CaSO4 + H2O(l) + CO2(g)
CaCO3 + H2SO4 = CaSO4 + H2O(l) + CO2(g)
1 mol 1 mol 1 mol 1 mol 1 mol
m (CO2) = 1 mol * 44g/mol = 44 g
m (H2O) = 1 mol * 18g/mol = 18 g
m (solution) = 44 g + 18 g = 62 g
% = (44 g / 66 g) * 100 = 70,96%
or
CaCO3 + H2SO4 = CaSO4 + H2O(l) + CO2(g)
n (CO2) = 1 dm3 / 22,4 dm3/mol = 0,045 mol
m (CO2) = 0,045 mol * 44 g/mol = 1,98 g
n (H2O) = 0,045 mol (1:1)
m (H2O) = 0,045 mol * 18 g/mol = 0,81 g
m (solution) 1,98 g + 0,81 g = 2,79 g
% = (1,98 g / 2,79 g) * 100 = 70,96%
Does it even make sense?
1. Homework Statement
CaCO3 + H2SO4 = CaSO4 + H2O(l) + CO2(g)
Homework Equations
The Attempt at a Solution
CaCO3 + H2SO4 = CaSO4 + H2O(l) + CO2(g)
1 mol 1 mol 1 mol 1 mol 1 mol
m (CO2) = 1 mol * 44g/mol = 44 g
m (H2O) = 1 mol * 18g/mol = 18 g
m (solution) = 44 g + 18 g = 62 g
% = (44 g / 66 g) * 100 = 70,96%
or
CaCO3 + H2SO4 = CaSO4 + H2O(l) + CO2(g)
n (CO2) = 1 dm3 / 22,4 dm3/mol = 0,045 mol
m (CO2) = 0,045 mol * 44 g/mol = 1,98 g
n (H2O) = 0,045 mol (1:1)
m (H2O) = 0,045 mol * 18 g/mol = 0,81 g
m (solution) 1,98 g + 0,81 g = 2,79 g
% = (1,98 g / 2,79 g) * 100 = 70,96%
Does it even make sense?
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