1)How many grams of silver chloride can be prepared by the reaction of 100.0 mL

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 ?

Explanation / Answer

Answer:

given reaction is-

AgNO3 (aq) + NaCl (aq) --------------> AgCl (s) + NaNO3 (aq)

from above reaction, we observed that, to producce 1 mole of AgCl(s) precipitation and 1 mole of NaNO3 (aq) we required 1 mole of AgNO3 (aq) and 1 mole of NaCl (aq) .

AgCl is precipitated, so ions remaining in solution (NaNO3 (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 NaNO3.

therefore moles of NaNO3 =   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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