If 3.000g of k2co3 were used in this experiment (instead of khco3), a. what is the balanced equation for the equation? k2co3 + 2hcl = 2kcl + co2 + h2o b. how many milliliters of 6.0 m hcl would be needed hcl c. how many grams of kcl would be formed in the reaction? g kcl

A. The balanced chemical equation is correct:

k2co3 + 2hcl --> 2kcl + co2 + h2o

B. First let us calculate the moles of K2CO3.

moles K2CO3 = 3 g / (138.205 g/mol) = 0.0217 mol

From the equation, we need 2 moles of HCl per 1 mole of K2CO3, hence:

moles HCl needed = 0.0217 mol * 2 = 0.0434 mol

So the volume required is:

volume HCl = 0.0434 mol / 6 M = 7.24 x 10^-3 L = 7.24 mL

C. The number of moles of HCl and KCl is equal hence:

moles KCl = 0.0434 mol

The molar mass of KCl is 74.5513 g/mol, so the mass is:

mass KCl = 0.0434 mol * 74.5513 g/mol = 3.24 grams