Consider a reaction similar to the precipitation reaction you will perform in th

Question

Consider a reaction similar to the precipitation reaction you will perform in the lab (hint: first balance the equation!): 25.0 mL of a 0.6 M CaCl2 solution was mixed with 35.0 mL of 1.4 M Na2CO3 solution to produce a precipitate of CaCO3. After drying, the mass of the precipitate obtained was 1.061 g. Calculate the percent yield (% yield) of CaCO3 obtained in the reaction. (Use 1 digit after the decimal place; do not use units, i.e. g, in your answer). Molar Masses (g/mol): Na2CO3 = 105.99 CaCl2= 110.98 CaCO3= 100.09

Explanation / Answer

CaCl2 + Na2CO3 --> CaCO3 + 2NaCl

Find limiting reactant

moles of CaCl2 = 0.6 M x 25 ml = 15 mmol

moles Na2CO3 = 1.4 M x 35 ml = 49 mmol

limiting reactant = CaCl2

theoretical yield of CaCO3 = 15 mmol x 100.09 g/mol/1000 = 1.50135 g

actual yield = 1.061 g

Percent yield of CaCO3 = 1.061 g x 100/1.50135 g = 70.7%

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