BaCl2(aq) + K2CrO4(aq)
BaCrO4(s) + 2 KCl(aq)
How many milliliters of 0.1200 M K2CrO4(aq) will react with 0.045 mol of BaCl2(aq)?
Use the coefficients in the balanced equation to convert moles BaCl2 to moles K2CrO4. Then M K2CrO4 = moles/L; you know M and moles, solve for L, then convert to mL.
To determine the number of milliliters of 0.1200 M K2CrO4(aq) that will react with 0.045 moles of BaCl2(aq), we need to use stoichiometry.
1. Write the balanced equation for the reaction:
BaCl2(aq) + K2CrO4(aq) -> BaCrO4(s) + 2 KCl(aq)
2. Determine the stoichiometric ratio between BaCl2(aq) and K2CrO4(aq) from the balanced equation:
According to the equation, 1 mole of BaCl2(aq) reacts with 1 mole of K2CrO4(aq).
3. Convert the moles of BaCl2(aq) to the required moles of K2CrO4(aq):
Since the stoichiometric ratio is 1:1, the number of moles of K2CrO4(aq) required is also 0.045 moles.
4. Use the molarity and volume relationship to find the volume of K2CrO4(aq):
Molarity (M) = moles/volume (L)
Rearranging the equation, volume (L) = moles/Molarity
Given: Molarity of K2CrO4(aq) = 0.1200 M
Moles of K2CrO4(aq) = 0.045 moles
Using the formula: volume (L) = moles/Molarity
volume (L) = 0.045 moles / 0.1200 M = 0.375 L
5. Convert the volume from liters to milliliters:
1 L = 1000 mL
volume (mL) = 0.375 L * 1000 mL/L = 375 mL
Therefore, 375 milliliters of 0.1200 M K2CrO4(aq) will react with 0.045 moles of BaCl2(aq).