1)
Balanced chemical equation is:
Ca(OH)2(aq) + 2 HCl(aq) ---> CaCl2(aq) + 2 H2O(l)
2)
Molar mass of Ca(OH)2,
MM = 1*MM(Ca) + 2*MM(O) + 2*MM(H)
= 1*40.08 + 2*16.0 + 2*1.008
= 74.096 g/mol
mass(Ca(OH)2)= 6.71 g
use:
number of mol of Ca(OH)2,
n = mass of Ca(OH)2/molar mass of Ca(OH)2
=(6.71 g)/(74.1 g/mol)
= 9.056*10^-2 mol
volume of HCl, V = 38.5 mL
= 3.85*10^-2 L
use:
number of mol in HCl,
n = Molarity * Volume
= 0.36*3.85*10^-2
= 1.386*10^-2 mol
1 mol of Ca(OH)2 reacts with 2 mol of HCl
for 9.056*10^-2 mol of Ca(OH)2, 0.1811 mol of HCl is required
But we have 1.386*10^-2 mol of HCl
so, HCl is limiting reagent
Answer: hydrochloric acid
3)
Molar mass of CaCl2,
MM = 1*MM(Ca) + 2*MM(Cl)
= 1*40.08 + 2*35.45
= 110.98 g/mol
According to balanced equation
mol of CaCl2 formed = (1/2)* moles of HCl
= (1/2)*1.386*10^-2
= 6.93*10^-3 mol
use:
mass of CaCl2 = number of mol * molar mass
= 6.93*10^-3*1.11*10^2
= 0.7691 g
Answer: 0.769 g
4)
According to balanced equation
mol of Ca(OH)2 reacted = (1/2)* moles of HCl
= (1/2)*1.386*10^-2
= 6.93*10^-3 mol
mol of Ca(OH)2 remaining = mol initially present - mol reacted
mol of Ca(OH)2 remaining = 9.056*10^-2 - 6.93*10^-3
mol of Ca(OH)2 remaining = 8.363*10^-2 mol
Molar mass of Ca(OH)2,
MM = 1*MM(Ca) + 2*MM(O) + 2*MM(H)
= 1*40.08 + 2*16.0 + 2*1.008
= 74.096 g/mol
use:
mass of Ca(OH)2,
m = number of mol * molar mass
= 8.363*10^-2 mol * 74.1 g/mol
= 6.197 g
Answer: 6.20 g
< Question 12 of 12 > A sample of 6.71 g of solid calcium hydroxide is...
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