You added 25 mL of a 0.03 mol/L KIO3 (potassium iodate) solution, about 1.5 g KI (potassium iodide, about 10 mmol) and 40 mL of 0.5 mol/L sulphuric acid (H2SO4) to your sample. The reaction proceeds according to the equation below:IO−3+5I−+6H+→3I2+3H2OIO3−+5𝐼−+6𝐻+→3𝐼2+3𝐻2𝑂 How much of KI is actually consumed by the reaction? 0.75 mmol 1.50 mmol 2.25 mmol 3.50 mmol 3.75 mmol
Question
You added 25 mL of a 0.03 mol/L KIO3 (potassium iodate) solution, about 1.5 g KI (potassium iodide, about 10 mmol) and 40 mL of 0.5 mol/L sulphuric acid (H2SO4) to your sample. The reaction proceeds according to the equation below:IO−3+5I−+6H+→3I2+3H2OIO3−+5𝐼−+6𝐻+→3𝐼2+3𝐻2𝑂 How much of KI is actually consumed by the reaction? 0.75 mmol 1.50 mmol 2.25 mmol 3.50 mmol 3.75 mmol
Solution
The reaction equation shows that one mole of KIO3 reacts with five moles of KI.
First, we need to find out how many moles of KIO3 we have.
Moles of KIO3 = Molarity * Volume(L) = 0.03 mol/L * 0.025 L = 0.00075 mol
Since one mole of KIO3 reacts with five moles of KI, the moles of KI that will react with 0.00075 mol of KIO3 will be 0.00075 mol * 5 = 0.00375 mol or 3.75 mmol.
So, the amount of KI actually consumed by the reaction is 3.75 mmol.
Similar Questions
You need to prepare 0.500 L of a solution of potassium sulfate.The required molar concentration (molarity) is 0.127 mol L-1 potassium sulfate.Calculate the mass of potassium sulfate in grams required for this solution.Enter your answer with three significant figures, do not include units.
Balance the following equations by adding the stoichiometric ratios and ionic charges (include the + or - charge). Do not leave the boxes blank or it will mark you incorrect. If the stoichiometry is 1 please insert 1 in the box(es) below. You don't have to add a 1 for the charge.Potassium iodide is dissolved in water KI(s) → K (aq) + I (aq)
The formula H2SO4 + 2 KOH --> K2SO4 + 2 H2O describes the reaction between sulfuric acid and potassium hydroxide. In the titration done, 20 mL of KOH with an unknown concentration is put into a flask with bromthymol blue indicator. The KOH solution was titrated with a 0.25 M H2SO4 solution. Three trials were performed and each trial yielded to an endpoint of yellow-colored solution, which is the desired endpoint. After three trials, the following readings were obtained: 15.5, 15.1, and 14.9 mL respectively.Calculate the equivalent moles of Potassium hydroxide used in the titration. (Answer must be rounded off to 4 decimal places)
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)
What mass of potassium sulfate is needed to prepare a 0.250 M potassium sulfate solution in a 500.0 mL volumetric flask? Express your answer to the correct number of significant figures.
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.