Answer:
Mass = 8.46 g
Explanation:
Given data:
Mass of water produced = ?
Mass of glucose = 20 g
Mass of oxygen = 15 g
Solution:
Chemical equation:
C₆H₁₂O₆ + 6O₂ → 6H₂O + 6CO₂
Number of moles of glucose:
Number of moles = mass/molar mass
Number of moles = 20 g/ 180.16 g/mol
Number of moles = 0.11 mol
Number of moles of oxygen:
Number of moles = mass/molar mass
Number of moles = 15 g/ 32 g/mol
Number of moles = 0.47 mol
now we will compare the moles of water with oxygen and glucose.
C₆H₁₂O₆ : H₂O
1 : 6
0.11 : 6/1×0.11 = 0.66
O₂ : H₂O
6 : 6
0.47 : 0.47
Less number of moles of water are produced by oxygen thus it will limit the yield of water and act as limiting reactant.
Mass of water produced:
Mass = number of moles × molar mass
Mass = 0.47 mol ×18 g/mol
Mass = 8.46 g
Answer:
The answer to your question is: CH₄ + 3/2 O₂ ⇒ CO₂ + 2 H₂O
Explanation:
Methane = CH₄
Oxygen = O
Carbon dioxide = CO₂
Water = H₂O
CH₄ + 3/2 O₂ ⇒ CO₂ + 2 H₂O
This is the balanced equation
Answer:
D. The presence of a base catalyst.
E. The presence of an acid catalyst. and
F. The presence of heat.
Explanation:
