How many milliliters of 6.0 M HCl would be needed if 3.000 g of K2CO3 were used in an experiment?

To determine the number of milliliters of 6.0 M HCl needed to react with 3.000 g of K2CO3, we can follow these steps:

Step 1: Calculate the number of moles of K2CO3.
To do this, we need to use the molar mass of K2CO3, which is calculated by adding up the atomic masses of each element in the compound. The atomic masses of K (potassium), C (carbon), and O (oxygen) can be found on the periodic table.
The molar mass of K2CO3 is:
2(K) + (C) + 3(O) = 2(39.10 g/mol) + 12.01 g/mol + 3(16.00 g/mol) = 138.21 g/mol

To calculate the number of moles of K2CO3, we can use the formula:
moles = mass / molar mass.

Therefore, moles of K2CO3 = 3.000 g / 138.21 g/mol.

Step 2: Write a balanced chemical equation for the reaction between K2CO3 and HCl.
The balanced equation is:
2K2CO3 + 2HCl → 2KCl + H2O + CO2

From the balanced equation, we can determine the stoichiometric ratio between K2CO3 and HCl. In this case, the ratio is 2:2, or 1:1. This means that one mole of K2CO3 reacts with one mole of HCl.

Step 3: Convert moles of K2CO3 to moles of HCl.
Since the stoichiometric ratio is 1:1, the number of moles of HCl required will be the same as the number of moles of K2CO3 calculated in Step 1.

Step 4: Calculate the volume of 6.0 M HCl needed.
Now, we need to use the concentration of the HCl solution, which is 6.0 M. This means there are 6.0 moles of HCl per liter of solution.

We can use the formula:
volume (L) = moles / concentration (M).

In this case, we need to convert the volume from liters to milliliters since the question asks for the answer in milliliters.

Therefore, volume (mL) = (moles of HCl / 6.0 M) × 1000 mL/L.

By following these steps, you should be able to find the number of milliliters of 6.0 M HCl required to react with 3.000 g of K2CO3.