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14 December, 12:49

For each reaction, calculate how many moles of the barium product you will produce using stoichiometry and the balanced reactions. since there are two reactants, calculate the moles of product using each reactant, and then use the number of moles which is less) this comes from the limiting reactant).

Moles of Ba (NH2SO3) 2.

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  1. 14 December, 16:37
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    given that

    mass of Ba (NO3) 2 = 1.40g

    mass of NH2SO3H = 2.50 g

    1) to determine the mole of Ba (NO3) 2

    2) to determine the mass of all three product formed in the reaction

    reaction

    Ba (NO3) 2 + 2NH2SO3H → Ba (NH2SO3) 2 + 2HNO3

    Solution

    we calculate the molar mass of each species by using their atomic masses

    BA = 137.33g/mol

    N = 14g/mol

    O = 16g/mol

    H = 1g/mol

    S = 32g/mol

    calculation

    Ba (NO3) 2 = Ba + 2N + 6O

    = 137.33 + 2X 14 + 6 X 16

    = 261.33g/mol

    NH2SO3H = N + 3H + S + 3O

    =14 + 3X1 + 32 + 3X 16

    = 97g/mol

    Ba (NH2SO3) 2 = Ba + 2N + 4H + 2S + 6O

    = 137.33 + 2 X 14 + 4 X1 + 2X32 + 6 X 16

    = 329.33g/mol

    HNO3 = H + n + 3O

    = 1 + 14 + 3 X 16

    = 63g/mol
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