what volume of 0.955 M HCl, in milliliters, is required to titrate 2.152g of Na2CO3 to the equivalent point?

Na2CO3(aq) + 2HCl(aq)
2NaCl(aq) + CO2(g) + H2O(1)

can this please be explained in simple step by steps? i am sick and i am having trouble understanding anything.

Oh no, I'm sorry to hear that you're sick! But don't worry, I'll try my best to explain it step-by-step in a simple way. Grab a tissue and let's get started!

Step 1: Convert the mass of Na2CO3 to moles.
To do this, we need to know the molar mass of Na2CO3, which is 106 g/mol.
Using the formula: Moles = Mass / Molar Mass
Moles of Na2CO3 = 2.152 g / 106 g/mol

Step 2: Determine the stoichiometry of the reaction.
From the balanced equation, we can see that 1 mole of Na2CO3 reacts with 2 moles of HCl.

Step 3: Calculate the moles of HCl required.
Since the ratio is 1:2, we can multiply the moles of Na2CO3 by 2 to find the moles of HCl.
Moles of HCl = (2.152 g / 106 g/mol) * 2

Step 4: Calculate the volume of 0.955 M HCl required.
Now we can use the molarity formula: Moles = Molarity * Volume
Since we know the moles of HCl from Step 3 and the molarity is given (0.955 M), we can solve for the volume.
Volume of HCl = Moles of HCl / Molarity

Step 5: Convert the volume to milliliters.
Since the question asks for the volume in milliliters, we need to convert the answer from Step 4 to milliliters (mL) if it's not already.

And voila! There you have it, the volume of 0.955 M HCl in milliliters that is required to titrate 2.152g of Na2CO3. I hope this helps, and I wish you a speedy recovery! If you have any more questions, feel free to ask

Sure, I'll guide you step-by-step through the calculation.

Step 1: Calculate the moles of Na2CO3
To find the moles of Na2CO3, we divide the given mass (2.152g) by its molar mass, which is the sum of the atomic masses of each element in the compound. The molar mass of Na2CO3 is calculated as follows:
2(Na) + (C) + 3(O) = 2(23.0 g/mol) + 12.01 g/mol + 3(16.00 g/mol) = 105.99 g/mol

Now, divide the given mass by the molar mass:
2.152g / 105.99 g/mol = 0.0203 mol

Step 2: Determine the stoichiometric ratio between Na2CO3 and HCl
From the balanced chemical equation, we can see that the stoichiometric ratio between Na2CO3 and HCl is 1:2. This means that 1 mole of Na2CO3 reacts with 2 moles of HCl.

Step 3: Calculate the moles of HCl required
Multiply the moles of Na2CO3 by the stoichiometric ratio:
0.0203 mol Na2CO3 × 2 mol HCl / 1 mol Na2CO3 = 0.0406 mol HCl

Step 4: Convert moles to volume (in milliliters) using the molarity
We are given the molarity (concentration) of HCl, which is 0.955 M. Molarity is defined as moles of solute per liter of solution. We can use this information to calculate the volume of HCl needed.

Rearrange the formula for molarity:
Molarity (M) = Moles of solute (mol) / Volume of solution (L)

We can rearrange this formula to solve for the volume:

Volume of solution (L) = Moles of solute (mol) / Molarity (M)

Substituting the values into the formula:
Volume of solution (L) = 0.0406 mol / 0.955 M = 0.0426 L

To convert the volume from liters to milliliters, multiply by 1000:
Volume of solution (mL) = 0.0426 L × 1000 mL/L = 42.6 mL

So, the volume of 0.955 M HCl required to titrate 2.152g of Na2CO3 to the equivalent point is 42.6 mL.

Sure, I'll help you understand the steps to solve this problem.

Step 1: Balance the chemical equation:
The balanced equation is: Na2CO3(aq) + 2HCl(aq) → 2NaCl(aq) + CO2(g) + H2O(l)

Step 2: Determine the moles of Na2CO3:
To find the moles of Na2CO3, divide the given mass of Na2CO3 (2.152g) by its molar mass (which is the sum of the atomic masses of all the atoms in the formula). The molar mass of Na2CO3 is approximately 105.99 g/mol.

Moles of Na2CO3 = (2.152g) / (105.99 g/mol)

Step 3: Determine the stoichiometry:
Based on the balanced equation, we know that 1 mole of Na2CO3 reacts with 2 moles of HCl. Therefore, the number of moles of HCl required will be twice the number of moles of Na2CO3 calculated in Step 2.

Moles of HCl = 2 * (moles of Na2CO3)

Step 4: Calculate the volume of HCl solution:
Now, we need to determine the volume of the 0.955 M HCl solution required to provide the moles of HCl calculated in the previous step. It can be obtained by using the formula:

Volume (in L) = (Moles of HCl) / (Molarity of HCl)

Note: Make sure the units are consistent. Since the molarity is given in M (molar) and we want the volume in mL, you'll need to convert mL to L by dividing the volume by 1000.

Volume (in mL) = (Moles of HCl) / (Molarity of HCl) * 1000

Substitute the calculated values into the formula to get the final answer.

1. moles Na2CO3 = M x L = ?

2. Convert moles Na2CO3 to moles HCl using the coefficients in the balanced equation. You should get ?moles Na2CO3 x (2 moles HCl/1 mole Na2CO3) = ?moles HCl.
3. M HCl = moles HCl/L HCl. You know moles HCl an M HCl, solve for L HCl and convert to mL.

42.5 mL