Answer:
Notice that the number of atoms of
K
and
Cl
are the same on both sides, but the numbers of
O
atoms are not. There are 3
O
atoms on the the left side and 2 on the right. 3 and 2 are factors of 6, so add coefficients so that there are 6
O
atoms on both sides.
2KClO
3
(
s
)
+ heat
→
KCl(s)
+
3O
2
(
g
)
Now the
K
and
Cl
atoms are not balanced. There are 2 of each on the left and 1 of each on the right. Add a coefficient of 2 in front of
KCl
.
2KClO
3
(
s
)
+ heat
→
2KCl(s)
+
3O
2
(
g
)
The equation is now balanced with 2
K
atoms,
Carbon dioxide (CO2)
Argon (Ar)
Hydrogen (H)
Helium (He)
Answer:
V = 43.95 L
Explanation:
Given data:
Mass of CH₄ decomposed = 15.63 g
Volume of H₂O produced at STP = ?
Solution:
Chemical equation:
CH₄ + 2O₂ → 2H₂O + CO₂
Number of moles of CH₄:
Number of moles = mass/molar mass
Number of moles = 15.63 g/ 16 g/mol
Number of moles = 0.98 mol
Now we will compare the moles of H₂O with CH₄.
CH₄ : H₂O
1 : 2
0.98 : 2×0.98 = 1.96 mol
Volume of hydrogen:
PV = nRT
1 atm × V = 1.96 mol × 0.0821 atm.L/mol.K × 273.15 K
V = 43.95atm.L / 1atm
V = 43.95 L
