50.05 g of caco3 reacts with excess amount of hcl to produce cacl2, h2o and co2. how many grams of hcl is consumed.?

50.05 g of caco3 reacts with excess amount of hcl to produce cacl2, h2o and co2. how many grams of hcl is consumed.

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  • 5 months ago

    First of all please write the chemical formula correctly. The balanced equation with the correct formula is given below. This is standard universal (world wide) practice.

    CaCO3 + 2HCl = CaCl2 + H2O + CO2

    Next note the molar ratios , 1:2::1:1:1

    Next calculate the Relative molecular mass (Mr) of CaCO3

    Ca x 1 = 40 x 1 = 40

    C x 1 = 12 1 = 12

    O x 3 = 16 x 3 = 48

    40 + 12 + 48 = 100

    Next Calculate the moles

    moles = mass(g) / Mr = 50.05 / 100 =0.5005 moles

    moles(CaCO3) = 0.5005 ( and is equivalent to one ratio)

    Moles (HCl) = 0.5005 x 2 = 1.001 moles ( and is equivalent to '2' ratios).

    Hence

    1.001 = mass(HCl) / (35.5 + 1)

    mass(HCl) = 1.001 x 36.5

    mass(HCl) = 36.5365 g ( The amount consumed).

  • 5 months ago

    2 HCl + CaCO3 = CaCl2 + H2O + CO2

    1 mole of the carbonate reacts with 2 moles of acid

    moles of CaCO3 = 50.05 g / 100.0869 g/mol  = 0.500 moles

    moles of acid consumed = 1.00 mole 

    mass of HCl consumed = 1.00 mole  * 36.4609 g/mol  

    = 36.46g ( corrected to 4 significant figures) 

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