70.0 g. The decomposition of 125 g CaCO3 produces 700 g CaO.
MM = 100.09 56.08
CaCO3 → CaO + CO2
Mass 125 g
a) Moles of CaCO3 = 125 g CaCO3 x (1 mol CaCO3/100.09 g CaCO3)
= 1.249 mol CaCO3
b) Moles of CaO = 1.249 mol CaCO3 x (1 mol CaO/1 mol CaCO3)
= 1.249 mol CaO
c) Mass of CaO = 1.249 mol CaO x (56.08 g CaO/1 mol CaO) = 70.0 g
Answer:
2H2S + 3O2 → 2SO2 + 2H2O
Explanation:
Step 1: Data given
Hydrogen sulfide = H2S
Oxygen = O2
sulfur dioxide = SO2
water = H2O
Step 2: The unbalanced equation
H2S + O2 → SO2 + H2O
Step 3: Balancing the equation
H2S + O2 → SO2 + H2O
On the left side we have 2x O (in O2) and on the right side we have 3x O (2x in SO2 and 1x in H2O). To balance the amount of O, we have to multiply O2 (on the left side) by 3 and SO2 and H2O on the right side by 3.
H2S + 3O2 → 2SO2 + 2H2O
On the right side we have 4x H and on the left side we have 2x H. To balance the amount of H, we have to multiply H2S by 2.
Now the equation is balanced.
2H2S + 3O2 → 2SO2 + 2H2O
Answer is: 20 ions, left side.
Unbalanced half reaction: C₅O₅²⁻(g) → CO₃²⁻(aq).
1) There are 5 carbon atoms on the left side of half reaction and 1 carbon atom on right, so first add coefficient 5 in fron of carbon dioxide to balance carbon atoms: C₅O₅²⁻(g) → 5CO₃²⁻(aq).
2) Because there are more oxygen atoms on the right, add OH⁻ ions on the left side of half reaction and water on the right: OH⁻(aq) + C₅O₅²⁻(g) → 5CO₃²⁻(aq) + H₂O(l).
3) Balance oxygen (25 atoms on boths side) and hydrogen (20 atoms on both side of half reaction) atoms:
Balanced half reaction: 20OH⁻(aq) + C₅O₅²⁻(g) → 5CO₃²⁻(aq) + 10H₂O(l).
Moving, growing, falling?
