While investigating mole ratios, a group of students collected data for the chemical reaction 2 NaOH(aq) + CaCl2(aq) —> 2 NaCl(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 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
To determine the amount of Ca(OH)2(s) produced, we first need to determine the limiting reactant in the reaction. To do this, we calculate the moles of each reactant using their respective molar masses:
1. Calculate the moles of NaOH:
Molar mass of NaOH = 23g (Na) + 16g (O) + 1g (H) = 40 g/mol
Moles of NaOH = 3.00 g / 40 g/mol = 0.075 mol
2. Calculate the moles of CaCl2:
Molar mass of CaCl2 = 40g (Ca) + 2(35.5g) (Cl) = 111 g/mol
Moles of CaCl2 = 2.10 g / 111 g/mol = 0.0189 mol
The balanced chemical equation shows that 2 moles of NaOH react with 1 mole of CaCl2. Therefore, the mole ratio between NaOH and CaCl2 is 2:1.
3. Determine the limiting reactant:
Since the mole ratio is 2:1, we need twice as many moles of NaOH as CaCl2. In this case, CaCl2 is the limiting reactant because there are fewer moles of CaCl2 than the 2:1 ratio requires.
4. Calculate the theoretical yield of Ca(OH)2:
From the balanced chemical equation, we can see that 1 mol of CaCl2 produces 1 mol of Ca(OH)2. Therefore, the moles of Ca(OH)2 produced will be equal to the moles of CaCl2:
Moles of Ca(OH)2 = 0.0189 mol
5. Convert moles of Ca(OH)2 to grams:
Molar mass of Ca(OH)2 = 40g (Ca) + 2(16g) (O) + 2(1g) (H) = 74 g/mol
Mass of Ca(OH)2 = 0.0189 mol x 74 g/mol = 1.40 g
Therefore, when 3.00 g of NaOH and 2.10 g of CaCl2 are combined, the amount of Ca(OH)2 produced is 1.40 g.