Chemistry, 29.01.2020 06:47 aylineorozco836

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?

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Answer from: rylei

The grams of AgCl are formed will be "0.269 g".

According to the question,

The moles of AgNO₃ will be:

= $Molarity\times Volume$

= $0.500\times 7.8\times 10^{-3}$

= $3.9\times 10^{-3} \ mole$

The moles of NH₄Cl will be:

= $Molarity\times Volume$

= $0.300\times 0.25\times 10^{-3}$

= $1.875\times 10^{-3} \ mole$

As per the reaction,

→ $1 \ mole \ NH_4 Cl \ produce = 1 \ mole \ AgCl$

$1.875\times 10^{-3} \ NH_4Cl \ will = 1\times 1.875\times 10^{-3}$

$= 1.875\times 10^{-3} \ mole \ AgCl$

hence,

The mass of AgCl will be:

= $moles\times \ molar \ mass$

= $1.873\times 10^{-3}\times 143.32 \ g/mol$

= $0.269 \ g$

Thus the above answer is appropriate.

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Answer from: noeminm105

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

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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 forme...

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