Help please..stoichometry

Question

How many grams of silver sulfide can be generated from the reaction of 3.94 g of silver nitrate with excess sodium sulfide

Answer 1

first, determine the equation.

AgNO3 + Na2S --> NaNO3 + Ag2S

then, balance the equation if needed. since this is an unbalanced equation, you obviously need to balance it.

2AgNO3 + Na2S --> 2NaNO3 + Ag2S

Now determine the compounds involved in the problem. For this, Ag2S and AgNO3 are needed.

there are 2 moles of silver nitrate and 1 mole of Ag2S. remember the coefficients in the equation are now considered as the number of moles.

Then equate the compounds needed.

2 moles of AgNO3 = 1 mole of Ag2S

since mole = (wt. of substance / mol.wt. of the substance)

2 (_wt. of AgNO3_)    =   1 (_wt. of Ag2S_)

   (mol.wt.of AgNO3)          (mol.wt. of Ag2S)

the wt. of AgNO3 is already given, 3.94g

the mol.wt. of AgNO3 can be computed.

Element = Atomic Weight * num. of atoms = mol. wt.

    Ag     =       108           *       1              =  108

    N       =       14            *        1             =+  14

    O      =        16            *        3             =  _48_

                           total mol. wt. of AgNO3 =   170g

we are finding for the wt. Ag2S, so represent it first with a variable.

Let wt. of Ag2S be x.

the mol.wt. of Ag2S can be computed.

Element = Atomic Weight * num. of atoms = mol. wt.

    Ag     =       108           *       2              =  216

    S       =       32            *        1             =+_32_

                           total mol. wt. of Ag2S =     248g

Substitute the values to our principal equation.

= 2 (_wt. of AgNO3_)    =   1 (_wt. of Ag2S_)

   (mol.wt.of AgNO3)          (mol.wt. of Ag2S)

= 2 (_3.94g_)    =   1 (___x___)

      (  170g  )            (   248g  )

=  (_7.88g_)    =   (___x___)

    (  170g  )          (   248g  )

cross multiply.

= (7.88g) (248g)   =   (170g)(x)

              

=(_1,954.24g^2_) = _(170g)(x)_

  (       170g       )       (170g)

= x = 11.49 g

11.49 g of Ag2S can be generated from the reaction.

Answer 2

2AgNO3 + Na2S  --> 2NaNO3 + Ag2S

As you said, this is just a stoichiometry problem.

(3.94g AgNO3) * (1molAgNO3/169.88 g AgNO3) *(1 mol Ag2S/2 mol AgNO3) * (247.8 g Ag2S/ 1 mol Ag2S)

= 2.87 g Ag2S

Answer 3

2AgNO3 + Na2S = Ag2S + 2NaNO3

n(Ag2S)=0.5*n(AgNO3)

           =0.5*(m/M)

           =0.5(3.94/169.87)

           =0.0116mol

m(Ag2S)=n*M

            =0.0116*247.81

           = 2.87g