Liquid sample a has a higher vapor pressure than liquid sample b (both in closed containers). what can be said about sample a co
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First, let's count mole of 10 g Calcium Carbonate
mole = Mass / Molecular Mass
Calcium Carbonate = CaCO₃
Molecular Mass = Ar Ca + Ar C + (3 x Ar O)
Molecular Mass = 40 + 12 + (3 x 16)
Molecular Mass = 100
next
Mole of CaCO₃ = 10 gram / 100
Mole of CaCO₃ = 0,1 mol
then equal the reaction equation first
CaCO₃ + 2 HCl ==> CaCl₂ + CO₂ + H₂O (Equal)
To count the mass of carbon dioxide that produced we must know the mole of CO₂ first
we can count by coefficient comparison
mole CO₂ = x mole CaCO₃
mole CO₂ = (1/1) x 0,1 mole
mole CO₂ = 0,1 mole
so
Mass of CO₂ = mole CO₂ x Molecular Mass of CO₂
Mass of CO₂ = 0,1 mole x (12 + (2 x 16))
Mass of CO₂ = 0,1 mole x 44
Mass of CO₂ = 4,4 g
so, mass of carbon dioxide that's produced by 10 g of calcium carbonate on reaction with chloride acid is 4,4 g.