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# Answer to Question #136004 in Inorganic Chemistry for Khan

Question #136004
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
1
2020-10-01T07:55:08-0400

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%

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