Sulfuric acid reacts with aluminum hydroxide by double replacement. a. if 30.0 g of sulfuric acid reacts with 25.0 g of aluminum hydro

Question

Sulfuric acid reacts with aluminum hydroxide by double replacement.
a. if 30.0 g of sulfuric acid reacts with 25.0 g of aluminum hydroxide, identify the limiting reactant.
b. determine the mass of excess reactant remaining.
c. determine the mass of each product formed. assume 100% yield.

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2022-04-09T00:53:41+00:00 1 Answer 0

Answers ( 1 )

  1. Cora
    0
    2022-04-09T00:55:33+00:00

    a.
    The balanced equation for thereaction between sulfuric acidand aluminium hydroxide is,
    n 3Hu2082SOu2084 + 2Al(OH)u2083
    u2192 Alu2082(SOu2084)u2083 + 6Hu2082O

    nnmass of Hu2082SOu2084= 30.0 g n
    Molar mass ofHu2082SOu2084
    = 98 g/mol
    nmoles of Hu2082SOu2084
    = 30.0 g /98g/mol= 0.306 mol

    nmass of Al(OH)u2083 = 25.0 g
    nMolar mass ofAl(OH)u2083
    = 78 g/mol
    nmoles of Al(OH)u2083 = 25.0 g/78 g/mol = 0.321 mol

    nStoichiometric ratio between Hu2082SOu2084 andAl(OH)u2083 is 3 : 2

    nHence reacted moles of Hu2082SOu2084 = 0.306 mol
    n reactednmoles ofAl(OH)u2083 =n0.306 mol x (2 / 3) = 0.204 mol

    Hence the limiting reactant is Hu2082SOu2084

    b.
    According to the above calculation, the excessnreactant is Al(OH)u2083.

    nThe reacted moles of Al(OH)u2083 =0.306 mol x (2 / 3) = 0.204 mol

    nThe added moles of Al(OH)u2083 = 0.321 mol

    nHence the remaining Al(OH)u2083 moles = added moles – reacted moles
    n n n =n0.321 mol – 0.204 mol
    n n n =n0.117 mol

    nMolar mass ofAl(OH)u2083= 78 g/mol
    Remaining massof Al(OH)u2083 =nnumber of moles x molar mass

    n n n = 0.117 mol x 78 g/mol
    n n n = 9.126 g
    n

    c.

    The products formed from the reaction betweennaluminium hydroxide and sulfuric acid are Alu2082(SOu2084)u2083 andHu2082O

    nThe limiting reactant is Hu2082SOu2084

    nThe stoichiometric ratio between Hu2082SOu2084nand Alu2082(SOu2084)u2083is 3 : 1
    nReacted moles of Hu2082SOu2084 =n0.306 mol

    nHence the moles of Alu2082(SOu2084)u2083 formed = 0.306 mol / 3

    n n n n = 0.102 mol
    nMolar mass of Alu2082(SOu2084)u2083
    = 342 g/mol
    nMass of Alu2082(SOu2084)u2083
    formed = 0.102 mol x 342 g/mol
    n n =34.884 g

    nThe stoichiometric ratio between Hu2082SOu2084nand Hu2082Ois 3 : 6
    nReacted moles of Hu2082SOu2084 =n0.306 mol

    nHence the moles of Hu2082Oformed = 0.306 mol x (6 /3)

    n n n = 0.612 mol
    nMolar mass of Hu2082O = 18 g/mol

    nMass of Hu2082O formed = 0.612mol x 18 g/mol

    n =11.016 g

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