1.00 g of alkali metal carbonate, M_2CO_3, is dissolved in water and made up to 250 cm^3 in a volumetric flask. 25.00 cm^3 portions of this solution are titrated against 0.113M hydrochloric acid. The average titre is 12.80cm^3. What element is M?
1.00 g of alkali metal carbonate, M_2CO_3 , is dissolved in water and made up to 250 cm^3 in a volumetric flask.
25.00 cm^3 portions of this solution are titrated against 0.113M hydrochloric acid. The average titre is 12.80cm^3
Calculate:
a. The moles of HCl in the titre
b. The moles of M_2CO_3 in 25 cm3 of solution
c. The moles of M_2CO_3 in 250 cm3 of solution
d. The relative formula mass of M_2CO_3
e. The identity of M
1.00 g of alkali metal carbonate,
Calculate:
a. The moles of HCl in the titre
b. The moles of
c. The moles of
d. The relative formula mass of
e. The identity of M
1 Answer
One necessary initial step is to write the stoichiometric equation....
Explanation:
We know that the formula mass of
Since the mass of
And a quick glance at the Periodic Table confirms that we gots potassium metal, and the carbonate is