Question

In the synthesis of barium carbonate from an alkali metal carbonate (M2CO3 where M is one of the alkali metals) a student generated 3.7 g of barium carbonate from 2.0 g of their alkali metal carbonate. M2CO3(aq) + BaCl2 (aq) --> 2MCl (aq) + BaCO3(s) How many moles of alkali metal carbonate were reacted? Question 3 options: 0.019 mol 0.038 mol 0.094 mol 2.3 mol

Answer 1

Answer:

0.019 moles of M2CO3

Explanation:

M2CO3(aq) + BaCl2 (aq) --> 2MCl (aq) + BaCO3(s)

From the equation above;

1 mol of  M2CO3 reacts to produce 1 mol of BaCO3

Mass of BaCO3 formed = 3.7g

Molar mass of BaCO3 = 197.34g/mol

Number of moles = Mass / Molar mass = 3.7 / 197.34 = 0.0187  ≈ 0.019mol

Since 1 mol of M2CO3 reacts with 1 mol of BaCO3,

1 = 1

x = 0.019

x = 0.019 moles of M2CO3