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 happen if an excess of NaOH was used. Their teacher approved a procedure in
which solutions containing an excess of NaOH and 2.10g of CaCl2 were combined.
Based on the quantities of these reactants, how many moles of
Ca(OH)2(s) is produced? Enter the value
The balanced chemical equation is 2 NaOH(aq) + CaCl2(aq) -> 2 NaCl(aq) + Ca(OH)2(s)
From the equation, we can determine the mole ratio of CaCl2 to Ca(OH)2 is 1:1. Therefore, the moles of Ca(OH)2 produced will be the same as the moles of CaCl2 used.
First, we need to calculate the number of moles of CaCl2:
Molar mass of CaCl2 = 40.08 g/mol + 2(35.45 g/mol) = 110.98 g/mol
Number of moles of CaCl2 = 2.10 g / 110.98 g/mol = 0.0189 moles
Therefore, 0.0189 moles of Ca(OH)2(s) is produced.