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)?
To determine the volume of the K2CrO4 solution that will react with 0.045 mol of BaCl2, we can use the concept of stoichiometry.
Given:
- Moles of BaCl2 = 0.045 mol
- Concentration of K2CrO4 = 0.1200 M
First, let's write the balanced equation for the reaction:
BaCl2(aq) + K2CrO4(aq) → BaCrO4(s) + 2 KCl(aq)
From the balanced equation, we can see that 1 mole of BaCl2 reacts with 1 mole of K2CrO4. Therefore, the number of moles of K2CrO4 needed to react with 0.045 mol of BaCl2 is also 0.045 mol.
Next, we'll use the formula:
Molarity (M) = Moles (mol) / Volume (L)
We rearrange the formula to solve for volume:
Volume (L) = Moles (mol) / Molarity (M)
Substituting the values into the formula, we have:
Volume (L) = 0.045 mol / 0.1200 M
Now, since the concentration is given in moles per liter (M), we need to convert the volume from liters to milliliters by multiplying by 1000:
Volume (mL) = 0.045 mol / 0.1200 M * 1000 mL/L
Calculating this expression, we find:
Volume (mL) = 375 mL
Therefore, 375 milliliters of 0.1200 M K2CrO4(aq) will react with 0.045 mol of BaCl2(aq).