From the reaction above, the rate is given by the following formula:
r = -(1/2) dA / dt = - dB / dt = (1/3) dC/ dt
Note that A and B charge is negative due to they decrease with time
given dA / dt = -0.110 M/s
hence dB / dt = -0.110 / 2 = -0.055 M/s
dC / dt = (-3/2) (-0.110) = 0.165 M/s
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
The salt is a solid compound and is considered the "Solute" of the solution.
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