0.125g of sodium dichromate is to be reacted wtih 0.025M potassium iodide. How many mLs of the potassium iodide solution will be needed.
1. balance equation is;
Na2Cr2O7 + 2KI --> 2NaI + K2Cr2O7
mass of Na2Cr2O7 = 0.125g
mole of Na2Cr2O7 = m/Mr = 0.125g/Mr =___mole
mole ratio
Na2Cr2O7 : KI
1 : 2
2.calculate the mole of KI using the mole ratio above and mole of Na2Cr2O7.
volume of KI = n/M where n is the mole you calculate in 2 for KI and M is the molarity of KI.
the molarity is in mol/L so the volume you will get is in liters.
hope that helps..
mole of
To solve this problem, we can use the principles of stoichiometry and the concept of molarity to find the number of milliliters (mL) of potassium iodide (KI) solution needed.
Step 1: Write the balanced chemical equation for the reaction between sodium dichromate (Na2Cr2O7) and potassium iodide (KI):
Na2Cr2O7 + 3KI -> 2NaI + Cr2O3 + KIO3
According to the balanced equation, the molar ratio between Na2Cr2O7 and KI is 1:3. This means that for every 1 mole of Na2Cr2O7, we need 3 moles of KI.
Step 2: Calculate the number of moles of Na2Cr2O7:
Given mass of Na2Cr2O7 = 0.125 g
Molar mass of Na2Cr2O7 = 2(23) + 2(52) + 7(16) = 294 g/mol
Number of moles of Na2Cr2O7 = Given mass / Molar mass
= 0.125 g / 294 g/mol
≈ 0.000425 moles
Step 3: Determine the number of moles of KI required:
From the balanced equation, we know that the molar ratio between Na2Cr2O7 and KI is 1:3. Therefore, the number of moles of KI required will be three times the moles of Na2Cr2O7.
Number of moles of KI = 3 * Number of moles of Na2Cr2O7
= 3 * 0.000425 moles
≈ 0.001275 moles
Step 4: Calculate the volume of the KI solution:
Given concentration of KI solution = 0.025 M (Molarity is measured in moles per liter, or mol/L)
Molarity (M) = Moles of solute / Volume of solution (in liters)
Rearranging the equation, Volume of solution (in liters) = Moles of solute / Molarity
Volume of KI solution (in liters) = 0.001275 moles / 0.025 M
= 0.051 L (since 1 L = 1000 mL)
Step 5: Convert the volume from liters to milliliters:
Volume of KI solution (in mL) = 0.051 L * 1000 mL/L
= 51 mL
Therefore, approximately 51 mL of the potassium iodide solution will be needed to react with 0.125 g of sodium dichromate.