The correct answer is reactants 2Na and 2H₂O and products 2NaOH and 1H₂.
The given unbalanced chemical equation is:
Na (s) + H2O (l) → NaOH (aq) + H₂ (g)
From the equation, it can be seen that there are three atoms of hydrogen on the products side, however, only two on the reactant's side. So, in order to balance the equation, one needs to multiply the sodium hydroxide by 2 to get a total of 4 atoms of hydrogen on the product's side.
This will enable one to readily double the number of molecules of water to get 4 atoms of hydrogen on the reactants side, and then balance the atoms of sodium by multiplying the sodium metal by 2. The balanced equation obtained is:
2Na (s) + 2H₂O (l) → 2NaOH (aq) + H₂ (g)
it is a polar covalent bond
<span>: The empirical formula for the compound is C3H60 (see below)
CO2 is the only product containing C,
C produced = 145.0 mg CO2 x (1 g / 1000 mg) x (1 mole CO2 / 44.0 g CO2) x (1 mole C / 1 mole CO2) = 0.00330 moles C.
H2O is the only product containing H,
H produced = 59.38 mg H2O x (1 g / 1000 mg) x (1 mole H2O / 18.0 g H2O) x (2 moles H / 1 mole H2O) = 0.00660 moles H.
Oxygen is in both and the unknown reacts with oxygen(in the air)
0.00330 moles C x (12.0 g C / 1 mole C) = 0.0396 g C = 39.6 mg C
0.00660 moles H x (1.01 g H / 1 mole H) = 0.00667 g H = 6.7 mg H
Because the unknown weighed 63.8 mg and consists off justC, H, and O, then
mass O = g unknown - g C - g H = 63.8 mg - 39.6 mg - 6.7 mg = 17.5 mg = 0.0175 g
0.0175 g O x (1 mole O / 16.0 g O) = 0.00109 moles O
The mole ratio of C:H:O is:
C = 0.00330
H = 0.00660
O = 0.00109
Divide by the smallest you get:
C = 0.00330 / 0.00109 = 3.03
H = 0.00660 / 0.00109 = 6.06
O = 0.00109 / 0.00109 = 1.00</span>
The answer is 15.5 gallons.
