Question

1)How many grams of silver chloride can be prepared by the reaction of 100.0 mL of 0.10 M silver nitrate with 100.0 mL of 0.20 M NaCl? Calculate the concentration of each ion remaining in solution after precipitation is complete.

2)How many grams of nitrogen dioxide is required to produce 7.50 g HNO₃.

3 NO₂ + H₂O --- 2HNO₃ + NO

3) 20.0 mL of 0.100 M NaOH is added to 40.0 mL of HCl of unknown concentration. Write a balanced reaction for the neutralization reaction and find out the concentration of HCl ?

Answer 1

given reaction is-

AgNO_{3 (aq)} + NaCl _(aq) --------------> AgCl (s) + NaNO_{3 (aq)}

from above reaction, we observed that, to producce 1 mole of AgCl(s) precipitation and 1 mole of NaNO_{3 (aq)} we required 1 mole of AgNO_{3 (aq)} and 1 mole of NaCl _(aq) .

AgCl is precipitated, so ions remaining in solution (NaNO_{3 (aq)}) is Na⁺ and NO₃^-.

we given that moles of  AgNO₃ = (100ml/1000) * .10 = .01 moles

& moles of NaCl = (100ml/1000) * 0.2 = 0.02 moles

( use molarity formula = moles of solute/ volume of solution in liter)

the limiting reagent in our reaction is AgNO₃ because it has fewer moles than NaCl, so the AgNO₃ will consume completely and a same number of moles will form of NaNO_3.

therefore moles of NaNO_{3 =}   moles of AgNO₃ comsumed = 0.01 moles

So, moles of Na⁺ ion = moles of NO₃^- = 0.01 ( after disociation )

volume of solution = 100 ml + 100 ml = 200ml

therefore cocerntration of  Na⁺ ion = 0.01/(200/1000) = cocerntration of NO₃^- = 0.05 M.

2) given that

mass of HNO₃ = 7.50 g

So, moles of HNO₃ = 7.50/ 64 = .1172 ( molar mass of HNO₃ = 64 g)

we know that Nitric acid is made by reaction of nitrogen dioxide (NO₂) with water.

3 NO₂ + H₂O ---> 2 HNO₃ + NO.

from above reaction, w observed that to produce 2 moles of HNO3, we required 3 moles of NO2 that means

for 1 moles of HNO3, we required 3/2 moles of NO2.

then to produce 0.1172 moles of HNO3

moles of NO2 required = (3/2)* 0.1172 = 0.1757 moles of NO2

3) given that

a volume (V1) of NaOH = 20ml, volume (V2) of HCl = 40 ml.

molarity of NaOH = 0.1 M, molarity of HCl = ?

balanced reaction is -

NaOH + HCl -----------------> NaCl + H₂O

we know that

M1V1 = M2V2 ( M1& M2 = molarity of NaOH andHCl )

So, 0.1 * 20ml = M2 * 40 ml

M2 = 0.05 M-- answer

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