How mucn 0.6 M barium chloride solution would be needed to completely react with 150mL of a 1.4 M sodium solution
Well, barium is ++ ion, and sodium is + ion, so it will thake half as much barium to react.
moles barium= 1/2*.150L*1.4M
or moles barium=.075*1.4
volume bariumchloride= .074*1.4/.6
I am poking my head in where I've not been asked to poke but I assume you intended to write sodium sulfate and not just sodium. I don't know how you react BaCl2 with sodium ion. Do they do anything except look at each other?
To determine how much 0.6 M barium chloride solution is needed to completely react with 150 mL of a 1.4 M sodium solution, we need to first identify the balanced equation for the reaction between barium chloride (BaCl2) and sodium (Na).
The balanced equation for the reaction is:
BaCl2 + 2Na → Ba + 2NaCl
From the equation, we can see that 1 mole of BaCl2 reacts with 2 moles of Na.
To calculate the amount of barium chloride needed, we can use the following steps:
Step 1: Convert the given volume of sodium solution to moles:
We have 150 mL of a 1.4 M sodium solution. To convert it to moles, we can use the formula:
moles = concentration (in M) x volume (in L)
Converting the given volume to liters: 150 mL ÷ 1000 = 0.15 L
Now, calculate the moles of sodium in 150 mL:
moles of Na = 1.4 M x 0.15 L = 0.21 moles of Na
Step 2: Determine the moles of barium chloride needed:
From the balanced equation, we know that 1 mole of BaCl2 reacts with 2 moles of Na. Therefore, we need half the moles of sodium for the reaction to be complete.
moles of BaCl2 needed = 0.5 x moles of Na = 0.5 x 0.21 = 0.105 moles of BaCl2
Step 3: Calculate the volume of 0.6 M barium chloride solution needed:
We know the concentration of the barium chloride solution is 0.6 M. To calculate the volume, we rearrange the formula:
volume (in L) = moles ÷ concentration (in M)
volume of BaCl2 = 0.105 moles ÷ 0.6 M = 0.175 L
Finally, we convert the volume from liters to milliliters to get the answer in the desired unit:
volume of BaCl2 = 0.175 L x 1000 = 175 mL
Therefore, 175 mL of 0.6 M barium chloride solution would be needed to completely react with 150 mL of a 1.4 M sodium solution.