Please solve question 5 throughly. The possible precipitates listed are the resu

Question

Please solve question 5 throughly. The possible precipitates listed are the resulting products from the three unbalanced chemical reactions listed below: _______ Ba(NO_3)_2 + _______ NH_2SO_3H + _______ H_2O rightarrow _______ Ba(NH_2SO_3)_2 + _______ HNO_3 _______ Ba(NO_3)_2 + _______ NH_2SO_3H + _______ H_2O rightarrow _______ BaSO_4 + ________ NH_4NO_3 + ________ HNO_3 _______ Ba(NO_3)_2 + _______ NH_2SO_3H + _______ H_2O rightarrow _______ Ba(NH_2)_2 + _______ H_2SO_4 + _______ HNO_3 Calculate the theoretical masses of the three possible products. Use the mass of your limiting reactant and the balanced chemical equations for the three possible reactions. Ba(NH_2SO_3)_2 BaSO_4 Ba(NH_2)_2

Explanation / Answer

Balanced chemical equation are

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

Ba(NO3)2 + NH2SO3H + H2O gives BaSO4 + NH4NO3 + HNO3

Ba(NO3)2 +2 NH2SO3H + 2H2O gives Ba (NH2)2 + 2 H2SO4 + 2 HNO3

Number of moles of limiting reagent barium nitrate = mass/molecular weight = 1.4509/ 261.33 = 0.00555 moles.

As per the balanced chemical equation

1 mole Ba(NO3)2 gives 1 mole Ba(NH2SO3) 2

1 mole Ba(NO3)2 gives 1 mole BaSO4

1 mole Ba(NO3)2 gives 1 mole Ba (NH2)2

Thus all products will be 1 mole equivalent of the reactant

Thus it will be 0.00555 moles.

Molar mass of Ba(NH2SO3) 2 is 329.4986 g/mol

Mass of this product = number of moles * molecular mas = 0. 00555*329.4986 = 1.8287 g (Ba(NH2SO3) 2 )

Molar mass of BaSO4is 233.38 g/mol

Mass of this product = number of moles * molecular mas = 0. 00555*233.38 =1.2952 g (BaSO4)

Molar mass of Ba (NH2)2is 169.3722 g/mol

Mass of this product = number of moles * molecular mas = 0. 00555*169.3722 =0.94 g (Ba (NH2)2)

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