When 7.80 ml of 0.500 m agno3 is added to 6.25 ml of 0.300 m nh4cl, how many grams of agcl are formed?

According to the question,

The moles of AgNO₃ will be:

=

=

=

The moles of NH₄Cl will be:

=

=

=

As per the reaction,

→

hence,

The mass of AgCl will be:

=

=

=

Thus the above answer is appropriate.

Learn more about solution here:

link

The answer to your question is 0.269 grams of AgCl

Explanation:

Data

[AgNO₃] = 0.50 M

Vol AgNO₃ = 7.80 ml

[NH₄Cl] = 0.30 M

Vol NH₄Cl = 6.25 ml

mass of AgCL

Balanced reaction

                 AgNO₃(aq)  +  NH₄Cl(aq)   ⇒   AgCl (s) + NH₄NO₃ (aq)

Process

1.- Calculate the moles of AgNO₃

Molarity = moles / volume

moles = Molarity x volume

moles = 0.50 x 0.0078

moles = 0.0039

2.- Calculate the moles of NH₄Cl

moles = 0.30 x 0.0063

moles = 0.00188

3.- Calculate the limiting reactant

The proportion of     AgNO₃(aq)  to  NH₄Cl(aq) is 1 :1, then, we conclude that the limiting reactant is NH₄Cl(aq), because there are less amount of this reactant in the experiment.

4.- Calculate the moles of AgCl

                     1 mol of NH₄Cl   1 mol of AgCl

              0.00188 mol of NH₄Cl x

                     x = (0.00188 x 1) /1

                     x = 0.00188 moles of AgCl

5.- Calculate the grams of AgCl

molecular mass of AgCl = 108 + 35.5 = 143.5 g

                         143.5 grams of AgCl 1 mol

                         x 0.00188 moles of AgCl

                          x = (0.00188 x 143.5) / 1

                          x = 0.269 grams of AgCl

its pwirhfpiwe

explanation:

because dgjpwojefopjwp