Question

I’ll mark as brainliest

Answer 1

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 = $\frac{coefficient \ of \ CO_2}{coefficient \ of \ CaCO_3}$  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.