Answer:
1) Mass = 18.864 g
2) V = 2.47 L
Explanation:
1) Given data:
Mass of water produced = ?
Mass of SnO₂ react = 79.0 g
Solution:
SnO₂ + 2H₂ → Sn + 2H₂O
Number of moles of SnO₂:
Number of moles = mass/molar mass
Number of moles = 79.0 g/ 150.71 g/mol
Number of moles = 0.524mol
Now we will compare the moles of SnO₂ and H₂O
SnO₂ : H₂O
1 : 2
0.524 : 2/1×0.524 =1.048 mol
Mass of water:
Mass = number of moles × molar mass
Mass = 1.048 mol × 18 g/mol
Mass = 18.864 g
2) Given data:
Volume of oxygen required = ?
Volume of C₃H₈ combust at STP = 0.499 L
Solution:
Chemical equation:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Number of moles of C₃H₈:
PV = nRT
R = general gas constant = 0.0821 atm.L /mol.K
1 atm × 0.499 L = n × 0.0821 atm.L /mol.K × 273.15 K
0.499 L atm.L = n × 22.43 atm.L /mol
n = 0.499 L atm.L / 22.43 atm.L /mol
n = 0.022 mol
Now we will compare the moles of propane and oxygen.
C₃H₈ : O₂
1 : 5
0.022 : 5/1×0.022 = 0.11 mol
Volume of oxygen required:
PV = nRT
R = general gas constant = 0.0821 atm.L /mol.K
1 atm × V = 0.11 mol × 0.0821 atm.L /mol.K × 273.15 K
1 atm × V = 2.47 atm.L
V = 2.47 atm.L / 1 atm
V = 2.47 L
