Answer:
Step-by-step explanation:
Given data:
Mass of CaCO₃ = 29.0 g
Mass of HCl = 11.0 g
Mass of CaCl₂ produced = ?
Which is excess reactant = ?
Mass remain unreacted = ?
Solution:
Chemical reaction:
CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
Number of moles of CaCO₃:
Number of moles = mass/ molar mass
Number of moles = 29.0 g/ 100.1 g/mol
Number of moles = 0.3 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 = 0.3 mol
Now we will compare the moles of HCl and CaCO₃ with CaCl₂.
HCl : CaCl₂
2 : 1
0.3 : 1/2×0.3 = 0.15 mol
CaCO₃ : CaCl₂
1 : 1
0.3 ; 0.3
The number of moles of CaCl₂ produced by HCl are less so it will limiting reactant.
Mass of CaCl₂:
Mass = number of moles × molar mass
Mass = 0.15 mol × 111 g/mol
Mass = 16.65 g
CaCO₃ is present in excess.
HCl : CaCO₃
2 : 1
0.3 : 1/2 ×0.3 = 0.15 mol
Mass remain unreacted:
Total moles of CaCO₃ = 0.3
Moles reacted = 0.15
Moles remain = 0.3 - 0.15 = 0.15 mol
Mass remain unreacted = 0.15 mol × molar mass
Mass remain unreacted = 0.15 mol × 100.1 g/mol
Mass remain unreacted = 15 g