Question

Calculate the mass of calcium carbonate produced if 15.2 mL of 0.306 M K2CO3 is reacted with 17.0 mL of 0.295 M CaCl2.

Answer 1

Answer:

$m_{CaCO_3}=0.465g$

Explanation:

Hello!

In this case, since the reaction between two aqueous solutions may turn out in the production of a solid precipitate, for potassium carbonate and calcium chloride, calcium carbonate is precipitated out as shown below:

$CaCl_2(aq)+K_2CO_3(aq)\rightarrow 2KCl(aq)+CaCO_3(s)$

Now, since the two reactants are in a 1:1 mole ratio, we infer they react in the same proportion, thus we compute the reacting moles, considering the used volumes of those molar solutions:

$n_{K_2CO_3}=0.0152L*0.306mol/L=0.00465mol\\\\n_{CaCl_2}=0.0170*0.295mol/L=0.00502mol$

Thus, since just 0.00465 mol out of 0.00502 moles of calcium chloride are consumed, the potassium carbonate is the limiting reactant, therefore the mass of yielded calcium carbonate (molar mass = 100.09 g/mol) is:

$m_{CaCO_3}=0.00465molK_2CO_3*\frac{1molCaCO_3}{1molK_2CO_3} *\frac{100.09gCaCO_3}{1molCaCO_3}\\\\m_{CaCO_3}=0.465g$

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