A FALSK CONTAINS 49.8 mL OF 0.150 CALCIUM HYDROXIDE. HOW MANY MILLILITERS OF 0.500 SODIUM CARBONATE ARE REQUIRED TO REACT COMPLETELY WITH THE CALCIUM HYDROXIDE IN THE FOLLOWING REACTION?

Na2CO3(aq)+ Ca(OH)2(aq)�¨CaCO3(s)+2NaOH(aq)

molesNa2CO3=molesCa(OH)2

MolarityNa2CO3(Volume)=.150*49.8ml
Volulme=.150*49.8/.5 ml

Well, this seems like a rather complicated chemical reaction. You know, I once tried to become a chemist, but I couldn't find the right mixture of elements... I guess you could say I had trouble "bonding" with the subject. But I digress!

To solve this problem, we need to use stoichiometry and mole ratios. We want to find out how many milliliters of 0.500M sodium carbonate are needed to react completely with 49.8 mL of 0.150M calcium hydroxide.

First, we need to convert the volume of calcium hydroxide to moles. We can use the formula M = moles/volume to do this. So, moles of calcium hydroxide = 0.150 x (49.8/1000).

Next, we need to determine the stoichiometric ratio between the coefficients of calcium hydroxide and sodium carbonate in the balanced equation. The balanced equation tells us that for every 1 mole of calcium hydroxide (Ca(OH)2), we need 1 mole of sodium carbonate (Na2CO3).

Finally, we can convert moles of calcium hydroxide to milliliters of sodium carbonate by using the formula volume = moles/M. So, volume of sodium carbonate = moles x (Molarity of sodium carbonate).

But remember, this is just the theoretical value. In reality, reactions may not always go to completion, so you might have some left over. Just like how we never really finish a bag of potato chips, there's always a few crumbs left at the bottom.

I hope that helps! Let me know if you have any other questions, chemical or otherwise.

To find how many milliliters of 0.500 M sodium carbonate are required to react completely with the calcium hydroxide, we need to use the stoichiometry of the balanced equation.

From the balanced equation:
1 mole of Ca(OH)2 reacts with 1 mole of Na2CO3.

To start the calculation, we need to determine the number of moles of Ca(OH)2 in the flask:

Moles of Ca(OH)2 = (volume of solution in liters) * (molarity of Ca(OH)2)

First, convert the volume of the solution from milliliters to liters:
49.8 mL = 49.8/1000 L = 0.0498 L

Next, calculate the moles of Ca(OH)2:
Moles of Ca(OH)2 = 0.0498 L * 0.150 mol/L = 0.00747 mol

Since the stoichiometry of the reaction is 1:1 between Ca(OH)2 and Na2CO3, the moles of Na2CO3 required will be equal to the moles of Ca(OH)2.

Therefore, we need 0.00747 mol of Na2CO3 to react completely with the calcium hydroxide.

Finally, to find the volume of the 0.500 M sodium carbonate solution needed, we can use the molarity formula:

Molarity = (moles of solute) / (volume of solution in liters)

Rearranging the formula, we can solve for the volume:

Volume of solution in liters = (moles of solute) / (molarity)

Substituting the values:
Volume of solution in liters = 0.00747 mol / 0.500 mol/L = 0.01494 L

Convert the volume to milliliters:
Volume of solution in milliliters = 0.01494 L * 1000 mL/L = 14.94 mL

Therefore, approximately 14.94 milliliters of 0.500 M sodium carbonate are required to react completely with 49.8 mL of 0.150 M calcium hydroxide.

To determine the number of milliliters of 0.500 M sodium carbonate required to react completely with 0.150 M calcium hydroxide, we first need to calculate the moles of calcium hydroxide present in the flask.

Step 1: Calculate the moles of calcium hydroxide (Ca(OH)2).
We can use the following formula to calculate the moles:

moles = volume (in liters) x molarity

Given:
Volume of calcium hydroxide solution = 49.8 mL = 0.0498 L
Molarity of calcium hydroxide solution = 0.150 M

moles of calcium hydroxide = 0.0498 L x 0.150 M = 0.00747 moles

Step 2: Use the balanced equation to determine the stoichiometric ratio between calcium hydroxide and sodium carbonate.
From the balanced equation: Na2CO3(aq) + Ca(OH)2(aq) → CaCO3(s) + 2NaOH(aq)
The stoichiometric ratio between calcium hydroxide and sodium carbonate is 1:1.

Step 3: Calculate the volume (in milliliters) of 0.500 M sodium carbonate required to react with 0.00747 moles of calcium hydroxide.

moles of sodium carbonate = moles of calcium hydroxide = 0.00747 moles

To convert moles to volume, use the following formula:

volume (in liters) = moles / molarity

Given:
Molarity of sodium carbonate solution = 0.500 M

volume of sodium carbonate = 0.00747 moles / 0.500 M = 0.0149 L

To convert liters to milliliters, multiply the value by 1000:

volume of sodium carbonate = 0.0149 L x 1000 mL/L = 14.9 mL

Therefore, 14.9 milliliters of 0.500 M sodium carbonate are required to react completely with 49.8 mL of 0.150 M calcium hydroxide.