Answer:
grams H₂O produced = 8.7 grams
Explanation:
Given 2C₂H₆(g) + 7O₂(g) => 4CO₂(g) + 6H₂O(l)
7g 18g ?g
Plan => Convert gms to moles => determine Limiting reactant => solve for moles water => convert moles water to grams water
Moles Reactants
moles C₂H₆ = 7g/30g/mol = 0.233mol
moles O₂ = 18g/32g/mol = 0.563mol
Limiting Reactant => (Test for Limiting Reactant) Divide mole value by respective coefficient of balanced equation; the smaller number is the limiting reactant.
moles C₂H₆/2 = 0.233/2 = 0.12
moles O₂/7 = 0.08
<u><em>Limiting Reactant is O₂</em></u>
Moles and Grams of H₂O:
Use Limiting Reactant moles (not division value) to calculate moles of H₂O.
moles H₂O = 6/7(moles O₂) = 6/7(0.562) moles H₂O = 0.482 mole H₂O yield
grams H₂O = (0.482mol)(18g·mol⁻¹) = 8.7 grams H₂O
the correct answer is.. ill tell you when i search it up gimme 2 seconds
Answer:
the boiling of 10g of liquid water
Starting in 1908, while a professor at the University of Chicago, Millikan worked on an oil-drop experiment in which he measured the charge on a single electron. J. J. Thomson had already discovered the charge-to-mass ratio of the electron.
