2C2H6 + 7O2 → 4CO2 + 6H2O
C2H6 is ethane,
Ethane reacts with oxygen, so it is the combustion.
The equation is balanced.
Answer is True.
Answer:
d. IF3
Explanation:
The Octet rule posits that atoms gain, atom lose, or share electrons in order to have a full valence shell of 8 electrons. This statement occurs when atoms also combine to form molecules until they attain or share eight valence electrons either by losing or gaining eletrons.
From the given options, a valid Lewis structure that cannot be drawn without violating the octet rule is IF3
Answer:
NaOH is the limiting reactant.
204.9 g of sodium phosphate are formed.
51.94 g of excess reactant will remain.
Explanation:
The reaction that takes place is:
- H₃PO₄ + 3NaOH → Na₃PO₄ + 3H₂O
First we <u>convert the mass of both reactants to moles</u>, using their <em>respective molar masses</em>:
- H₃PO₄ ⇒ 175 g ÷ 98 g/mol = 1.78 mol
- NaOH ⇒ 150 g ÷ 40 g/mol = 3.75 mol
1.78 moles of H₃PO₄ would react completely with (1.78 * 3) 5.34 moles of NaOH. There are not as many NaOH moles so NaOH is the limiting reactant.
--
We <u>calculate the produced moles of Na₃PO₄</u> using the <em>limiting reactant</em>:
- 3.75 mol NaOH *
= 1.25 mol Na₃PO₄
Then we <u>convert moles into grams</u>:
- 1.25 mol Na₃PO₄ * 163.94 g/mol = 204.9 g
--
We calculate how many H₃PO₄ moles would react with 3.75 NaOH moles:
- 3.75 mol NaOH *
= 1.25 mol H₃PO₄
We substract that amount from the original amount:
- 1.78 - 1.25 = 0.53 mol H₃PO₄
Finally we <u>convert those remaining moles to grams</u>:
- 0.53 mol H₃PO₄ * 98 g/mol = 51.94 g
When P1/P2 = C1/C2
and C is the molarity which = moles/volume
so, P1/P2 = [(mass1/mw)/volume] / [(mass2/mw)/volume]
P1/P2 = (mass1/mw)/1.5L / (mass2/mw)/1.5L
so, Mw and 1.5 L will cancel out:
∴P1/P2 = mass1 / mass2
∴ mass 2 = mass1*(P2 / P1)
= 0.278g * (78 bar / 62 bar)
= 0.35 g
∴ the quantity of argon that will dissolve at 78 bar = 0.35 g