Calculate the volume (in cm3) of oxygen (measured at STP) produced when 25.0 cm3 of 0.0180 mol dm-3 potassium manganate(VII) reacts with 30.0 cm3 0.0250 cm3 hydrogen peroxide solution in the presence of excess sulfuric acid, according to the equation:2KMnO4(aq) + 5H2O2(aq) + 3H2SO4(aq) → 2MnSO4(aq) + K2SO4(aq) + 5O2(g) + 8H2O(l)
Question
Calculate the volume (in cm3) of oxygen (measured at STP) produced when 25.0 cm3 of 0.0180 mol dm-3 potassium manganate(VII) reacts with 30.0 cm3 0.0250 cm3 hydrogen peroxide solution in the presence of excess sulfuric acid, according to the equation:2KMnO4(aq) + 5H2O2(aq) + 3H2SO4(aq) → 2MnSO4(aq) + K2SO4(aq) + 5O2(g) + 8H2O(l)
Solution
To solve this problem, we need to follow these steps:
Step 1: Calculate the number of moles of potassium manganate(VII) and hydrogen peroxide.
The number of moles = concentration x volume (in dm^3).
For potassium manganate(VII): Number of moles = 0.0180 mol/dm^3 x 25.0 cm^3 x (1 dm^3/1000 cm^3) = 0.00045 mol
For hydrogen peroxide: Number of moles = 0.0250 mol/dm^3 x 30.0 cm^3 x (1 dm^3/1000 cm^3) = 0.00075 mol
Step 2: Determine the limiting reactant.
From the balanced chemical equation, we can see that 2 moles of KMnO4 react with 5 moles of H2O2. So, the ratio of KMnO4 to H2O2 should be 2:5.
However, the ratio of the number of moles we have is 0.00045:0.00075, which simplifies to 3:5. This means we have more KMnO4 than required, so H2O2 is the limiting reactant.
Step 3: Calculate the number of moles of oxygen produced.
From the balanced chemical equation, we can see that 5 moles of H2O2 produce 5 moles of O2. So, the number of moles of O2 produced is equal to the number of moles of H2O2.
Therefore, the number of moles of O2 produced = 0.00075 mol.
Step 4: Calculate the volume of oxygen produced at STP.
At STP, 1 mole of any gas occupies 22.4 dm^3 (or 22400 cm^3).
Therefore, the volume of O2 produced = 0.00075 mol x 22400 cm^3/mol = 16.8 cm^3.
So, the volume of oxygen produced when 25.0 cm^3 of 0.0180 mol dm^-3 potassium manganate(VII) reacts with 30.0 cm^3 0.0250 cm^3 hydrogen peroxide solution in the presence of excess sulfuric acid is 16.8 cm^3.
Similar Questions
Calculate the volume of oxygen gas produced at 31.65 atm and 55.78 oC by the complete decomposition of 30.23 g of potassium chlorate. The balanced equation for this reaction is …2KClO3 ⟶ 2KCl + 3O2
1.MnO4-(aq) + 8H+(aq) + 5Fe2+(aq) ---------> Mn2+(aq) + 4H2O(l) + 5Fe3+(aq)2. Calculate the volume of iron (ii) sulfate into dm^3 (25.0 cm^3= 0.025dm^33.Calculate the number of moles for the potassium permanganate, (volume of Pottasium permangnate used) x (concentration of Potassium permanganate) = (Number of moles of Potassium Permanganate)4. Mole RatioMnO : Fe 1 : 5 5.Repeat for accuracy and calculate the average mass of iron per tablet from multiple measurements
The concentrations of solutions of hydrogen peroxide are often represented in terms of ‘volumestrength’. In this experiment you will determine the ‘volume strength’ of a solution of hydrogen peroxidebytitrationwithacidifiedpotassiummanganate(VII).2MnO 4–(aq)+5H2 O2(aq)+6H+(aq) 2Mn2+ (aq)+8H2O(l)+5O2(g)FA 1 is 0.0300 mol dm –3potassiummanganate(VII), KMnO4 .FA 2isdilutesulfuricacid,H2 SO4 .FA 3isaqueoushydrogenperoxide,H2 O2 .(a) MethodDilution of FA 3● Pipette25.0cm3 of FA 3intothe250cm3 volumetric flask.● Make the solution up to the mark using distilled water.● Shake the flask thoroughly.● Label this diluted solution of hydrogen peroxide FA 4.Titration● Fill the burette with FA 1.● Rinse the pipette thoroughly with distilled water and then with a little FA 4.● Pipette25.0cm3 of FA 4 into a conical flask.● Usethe25cm3 measuring cylinder to add 20 cm3 of FA 2 into the same conical flask.● Perform a rough titration and record your burette readings in the space below.The rough titre is .............................. cm3 .● Carry out as many accurate titrations as you think necessary to obtain consistent results.● Make sure any recorded results show the precision of your practical work.● Record all of your burette readings and the volume of FA 1 added in each accurate titration.Keep FA 1, FA 2 and FA 3 for use in Questions 2 and 3.[7]IIIIIIIVVVIVII39701/33/F/M/20©UCLES 2020 [Turn over(b) From your accurate titration results, obtain a suitable value for the volume of FA 1 to be usedin your calculations.Show clearly how you obtained this value. 25.0cm3 of FA 4required..............................cm3 of FA 1. [1](c) Calculations(i) Give your answers to (ii), (iii), (iv) and (v)totheappropriatenumberofsignificantfigures.[1](ii) Calculate the number of moles of potassium manganate(VII) present in the volumecalculated in (b).moles of KMnO 4 = .............................. mol [1](iii) The equation for the reaction of potassium manganate(VII) with hydrogen peroxide isshown.2MnO 4–(aq)+5H2 O2(aq)+6H+(aq) 2Mn2+ (aq)+8H2O(l)+5O2(g)Use your answer to (c)(ii) to calculate the number of moles of hydrogen peroxide used ineach titration. molesofH2 O2 = .............................. mol HencecalculatetheconcentrationofH2 O2 in FA 4, in mol dm –3 . concentrationofH2 O2 in FA 4 = .............................. mol dm–3
The equation for the reaction in stage 3 is shown.4N*O_{2} + 2H_{2}*O + O_{2} -> 4HNOCalculate the volume of O, gas, at room temperature and pressure ( r .t.p.) . needed to produce 1260g of HNO3. Use the following steps.11:55 AMCalculate the number of moles of HNO3.moles of HNO3 =Deduce the number of moles of O_{2} that reacted.moles of O_{2} =Calculate the volume of O2 gas that reacts at room temperature and pressure (r.t.p.).volume of O2 gas =dm³ [4]
Oxygen can be generated from the thermal decomposition of potassium chlorate, by the following reaction:2 KClO3(s) à 2 KCl(s) + 3 O2(g)What is the volume of oxygen produced, in litres (to 1 decimal place, input value only), at 173.9 kPa and 36.6 °C, when 27.4 grams of potassium chlorate are decomposed
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.