A mixture consists of lithium carbonate (Li 2 CO 3 ) and potassium carbonate (K

Question

A mixture consists of lithium carbonate (Li2CO3) and potassium carbonate (K2CO3). These react with hydrochloric acid as follows.

When 5.614 g of this mixture was analyzed, it consumed 75.14 mL of 1.770 M HCl.

Calculate the number of grams of lithium carbonate in the original mixture. (answer in grams)

Calculate the number of grams of potassium carbonate in the original mixture. (answer in grams)

Calculate the percentage of lithium carbonate in the original mixture. (answer in %)

Calculate the percentage of potassium carbonate in the original mixture. (answer in %)

A mixture consists of lithium carbonate (Li2CO3) and potassium carbonate (K2CO3). These react with hydrochloric acid as follows. When 5.614 g of this mixture was analyzed, it consumed 75.14 mL of 1.770 M HCl. Calculate the number of grams of lithium carbonate in the original mixture. (answer in grams) Calculate the number of grams of potassium carbonate in the original mixture. (answer in grams) Calculate the percentage of lithium carbonate in the original mixture. (answer in %) Calculate the percentage of potassium carbonate in the original mixture. (answer in %)

Explanation / Answer

Let there be 'x' g of Li2CO3 & (5.614-x) g of K2CO3

Molar mass of Li2CO3 = 73.891 g/mole

Molar mass of K2CO3 = 138.205 g/mole

Thus, moles of x g of Li2CO3 = x/73.891

Moles of (x-5.614)g of K2CO3 = (5.614-x)/138.205

Now, moles of HCl consumed = 1.77*0.07514 = 0.133

Thus, moles of HCl = 2*[moles of Li2CO3 + Molesof K2CO3]

or, 0.133 = [x/73.891] + [(5.614-x)/138.205]

or, 1358.2 = 2*[414.82 + 64.314x]

or, x = 4.11 = mass of  Li2CO3

mass of  K2CO3 = 1.504 g

% of  Li2CO3 in the mixture = (4.11/5.614)*100 = 73.21 %

% of  K2CO3 in the mixture = 100 - 73.21 = 26.79%

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