Question

Magnesium chloride can be made by reacting excess magnesium carbonate with dilute hydrochloric acid.
The equation for the reaction is
MgCO3 + 2HCl → MgCl2 + H2O + CO2
(a) In an experiment 0.050 mol of MgCO3 reacts with excess HCl.
A yield of 5.5 g of MgCl2.6H2O is obtained.
(i) Calculate the percentage yield of MgCl2.6H2O

Answer 1

STEP I: CALCULATE MASS OF MgCl₂ IN THE 5.5g OF THE PRODUCT

(Use RAM: Mg = 24.3, Cl = 35.5, O = 16.0, H = 1.0)

RFM of MgCl₂.6H₂O = 24.3 + (2 x 35.5) + (6x18)

= 203.3g

(We shall ignore the water of crystallization to make our work less complex)

Actual mass of MgCl₂ in the 5.5g of MgCl₂.6H₂O

5.5g $\to$ 203.3g

? $\to$ 95.3g

$=\dfrac{5.5gx95.3g}{203.3g} = 2.578g$ of MgCl₂

STEP II: CALCULATE THE THEORETICAL MASS OF MgCl₂ PRODUCED

From the equation, the mole ratios of MgCO₃ to MgCl₂ is 1:1

Thus moles of MgCl₃ produced is $= \dfrac{1}{1} x 0.050mol = 0.050moles$

Mass of MgCl3 from 0.050 moles and RFM (24.3) + (2x35.5) = 95.3g/mol

Mass = Moles x RFM

= 0.050mol x 95.3g/mol

= 4.765g

STEP III: CALCULATE % YIELD

Theoretical mass = 4.765g

Actual mass = 2.578g

% Yield $= \dfrac{Actual yield}{Theoretical yield} x 100$

$= \dfrac{2.578g}{4.765g} x 100$

= 54.10%