A + 2B + 3C rightleftharpoons AB_2C_3. Reacti | vedclass

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(A) The balanced chemical equation is: $A + 2B + 3C \rightleftharpoons AB_2C_3$ Calculate the moles of reactants: $n_A = \frac{6.0 \ g}{60 \ g/mol} =

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(A) The balanced chemical equation is: $A + 2B + 3C ightleftharpoons AB_2C_3$ Calculate the moles of reactants: $n_A = \frac{6.0 \ g}{60 \ g/mol} = 0.1 \ mol$ $n_B = \frac{6.0 	imes 10^{23}}{6 	imes 10^{23}} = 1.0 \ mol$ $n_C = 0.036 \ mol$ Determine the limiting reagent: For $A$: $0.1 / 1 = 0.1$ For $B$: $1.0 / 2 = 0.5$ For $C$: $0.036 / 3 = 0.012$ Since $C$ has the lowest ratio,it is the limiting reagent. Calculate moles of product $AB_2C_3$ formed: $n_{AB_2C_3} = \frac{n_C}{3} = \frac{0.036}{3} = 0.012 \ mol$ Calculate the molar mass of $AB_2C_3$: $M.M. = \frac{	ext{mass}}{	ext{moles}} = \frac{4.8 \ g}{0.012 \ mol} = 400 \ g/mol$ Calculate atomic mass of $B$ $(x)$: $400 = (1 	imes 60) + (2 	imes x) + (3 	imes 80)$ $400 = 60 + 2x + 240$ $400 = 300 + 2x$ $2x = 100$ $x = 50 \ amu$

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