Answer:
Explanation:
what wheres the answer???
Answer:
There is 52.33 grams of water produced.
Explanation:
Step 1: Data given
Mass of propane burned = 32.00 grams
Molar mass of propane = 44.1 g/mol
Oxygen is in excess
Molar mass of water = 18.02 g/mol
Step 2: The balanced equation
C3H8 + 5O2 → 4H2O + 3CO2
Step 3: Calculate moles of propane
Moles of propane = mass propane / molar mass of propane
Moles of propane = 32.00 grams / 44.1 g/mol
Moles of propane = 0.726 moles
Step 4: Calculate moles of H2O
Propane is the limiting reactant.
For 1 mol of propane consumed, we need 5 moles of O2 to produce 4 moles of H2O and 3 moles of CO2
For 0.726 moles of propane we'll have 4*0.726 = 2.904 moles of H2O
Step 5: Calculate mass of H2O
Mass of H2O = moles of H2O * molar mass of H2O
Mass of H2O = 2.904 moles * 18.02 g/mol
Mass of H2O = 52.33 grams
There is 52.33 grams of water produced.
Answer:
Explanation:
Discussion
When Pressure increases equilibrium shifts to the side with the smallest number of moles. But which side is that?
N2(g) + 3H2(g) ⇌ 2NH3(g)
The left side has 1 mol of nitrogen (N2) and 3 moles of Hydrogen = 4 mols
on the left side.
The right side has 2 mols of NH3 = 2 mols on the right.
Conclusion: You tell the number of mols by the Balance numbers to the left of each chemical in an equation.
Since the left side N2 + 3H2 = 4 mols, the equilibrium does NOT shift left.
2NH3 is only two mols.
The equilibrium shifts Right
Answer
D
The answer is B or the second answer