Student Question

While investigating mole ratios, a group of students collected data for the chemical reaction 2 NaOH(aq) + CaCl2 (aq) -> 2NaCl(aq) +Ca(OH)2 (s). The students wondered what would happened if an excess of NaOH was used. Their teacher approved a procedure in which solutions containing 3.00 g of NaOH and 2.10 g of CaCL2 were combined. Based on the quantities of these reactants, what amount of Ca(OH)2(s) is produced?

Expert Answers

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The balanced chemical reaction is as follows:

2 NaOH (aq) + CaCl2 (aq) -> 2 NaCl (aq) + Ca(OH)2 (s)

According to this chemical reaction, 2 moles of sodium hydroxide (NaOH) will react with 1 mole of calcium chloride (CaCl2) to produce 2 moles of sodium chloride (NaCl) and a mole of calcium hydroxide (Ca(OH)2).

When a substance is in excess quantity (which means it is present in a higher quantity than can be used in the chemical reaction), only the stoichiometric amount will be used, and the rest will be left over after the completion of the reaction.

To calculate if a substance is in excess in the given scenario, we will need to calculate the moles of the reactants and then check against the stoichiometric quantities needed for the reaction. The molecular masses of sodium hydroxide and calcium chloride are 40 gm/mole and 111 gm/mole, respectively.

3 gm of NaOH is given, which is equal to 3 gm / (40 gm/mole) = 0.075 moles of NaOH.

Similarly, the 2.1 gm of CaCl2 is equivalent to 2.1 gm/ (111 gm/mole) = 0.019 moles of CaCl2.

According to stoichiometry, 2 moles of NaOH will react with 1 mole of CaCl2. In other words, 0.019 moles of CaCl2 will react with 0.038 moles (= 2 x 0.019 moles) or 1.52 gm of NaOH. The leftover or excess amount of NaOH is equal to 0.037 moles (= 1.48 gm). In other words, about half the amount of NaOH will be left over after the reaction.

Each mole of CaCl2 that undergoes the given reaction produces one mole of Ca(OH)2. Since 0.019 moles of CaCl2 react, we can expect 0.019 moles of Ca(OH)2 as the product.

The molar mass of Ca(OH)2 is 74 gm/mole. In other words, about 1.4 gm (= 0.019 moles x 74 gm/moles) of calcium hydroxide will be produced at the completion of this reaction.

Similarly, we can expect 0.038 moles or 2.223 gm of NaCl as a product of this reaction.

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