Answer:
We need 17.2 L of Ca(OH)2
Explanation:
Step 1: Data given
Concentration of Ca(OH)2 = 1.45 M
Moles of H2SO4 = 25.0 moles
Step 2: The balanced equation
Ca(OH)2 + H2SO4 ⟶2H2O + CaSO4
Step 3: Calculate moles Ca(OH)2
For 1 mol Ca(OH)2 we need 1 mol H2SO4 to produce 2 moles H2O and 1 mol CaSO4
For 25.0 moles H2SO4 we'll need 25.0 moles Ca(OH)2 to produce 50 moles H2O and 25.0 moles CaSO4
Step 4: Calculate volume of Ca(OH)2
Volume Ca(OH)2 = moles Ca(OH)2 / concentration Ca(OH)2
Volume Ca(OH)2 = 25.0 moles / 1.45 M
Volume Ca(OH)2 = 17.2 L
We need 17.2 L of Ca(OH)2
Answer:
Mass = 14.3 g
Explanation:
Given data:
Mass of Mg(OH)₂ = 16.0 g
Mass of HCl = 11.0 g
Mass of MgCl₂ = ?
Solution:
Chemical equation:
Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O
Number of moles of Mg(OH)₂ :
Number of moles = mass/ molar mass
Number of moles = 16.0 g/ 58.3 g/mol
Number of moles = 0.274 mol
Number of moles of HCl :
Number of moles = mass/ molar mass
Number of moles = 11.0 g/ 36.5 g/mol
Number of moles = 0.301 mol
Now we will compare the moles of Mg(OH)₂ and HCl with MgCl₂.
Mg(OH)₂ : MgCl₂
1 : 1
0.274 : 0.274
HCl : MgCl₂
2 : 1
0.301 : 1/2×0.301 = 0.150
The number of moles of MgCl₂ produced by HCl are less so it will limiting reactant.
Mass of MgCl₂:
Mass = number of moles × molar mass
Mass = 0.150 × 95 g/mol
Mass = 14.3 g
Answer:
The calorimeter constant is = 447 J/°C
Explanation:
The heat absorbed or released (Q) by water can be calculated with the following expression:
Q = c × m × ΔT
where,
c is the specific heat
m is the mass
ΔT is the change in temperature
The water that is initially in the calorimeter (w₁) absorbs heat while the water that is added (w₂) later releases heat. The calorimeter also absorbs heat.
The heat absorbed by the calorimeter (Q) can be calculated with the following expression:
Q = C × ΔT
where,
C is the calorimeter constant
The density of water is 1.00 g/mL so 50.0 mL = 50.0 g. The sum of the heat absorbed and the heat released is equal to zero (conservation of energy).
Qabs + Qrel = 0
Qabs = - Qrel
Qcal + Qw₁ = - Qw₂
Qcal = - (Qw₂ + Qw₁)
Ccal . ΔTcal = - (cw . mw₁ . ΔTw₁ + cw . mw₂ . ΔTw₂)
Ccal . (30.31°C - 22.6°C) = - [(4.184 J/g.°C) × 50.0 g × (30.31°C - 22.6°C) + (4.184 J/g.°C) × 50.0 g × (30.31°C - 54.5°C)]
Ccal = 447 J/°C
Yes that’s why we see it in different shapes all the time
