Answer:
20 molecules of oxygen gas remains after the reaction.
Explanation:
Molecules of ethyne = 52
Molecules of oxygen gas = 150
According to reaction, 2 molecules of ethyne reacts with 5 molecules of oxygen gas.
Then 52 molecules of ethyne will react with:
of oxygen gas.
As we can see that we have 150 molecules of oxygen gas, but 52 molecules of ethyne will react with 130 molecules of oxygen gas. So, this means that ethyne is a limiting reagent and oxygen gas is an excessive reagent.
Remaining molecules of recessive reagent = 150 - 130 = 20
20 molecules of oxygen gas remains after the reaction.
<span>Let's assume
that the oxygen gas has ideal gas behavior.
Then we can use ideal gas formula,
PV = nRT</span>
Where, P is the pressure of the gas (Pa), V is the volume of the gas
(m³), n is the number of moles of gas (mol), R is the universal gas
constant ( 8.314 J mol⁻¹ K⁻¹) and T is temperature in Kelvin.
<span>
P = 2.2 atm = 222915 Pa
V = 21 L = 21 x 10</span>⁻³ m³
n = ?
R = 8.314 J mol⁻¹ K⁻¹
<span>
T = 87 °C = 360 K
By substitution,
</span>222915 Pa x 21 x 10⁻³ m³ = n x 8.314 J mol⁻¹ K⁻<span>¹ x 360 K
n
= 1.56</span><span> mol</span>
<span>
Hence, 1.56 moles of the oxygen gas are </span><span>
left for you to breath.</span><span>
</span>
The net equations are obtained from the double displacement of the cations and anions, then balance.
NH3(aq) + HC2H3O2 (aq) = NH4+(aq) + C2H3O2-(aq<span>)
</span><span>H+(aq) + C2H3O2-(aq) + NH3(aq) -> NH4+(aq) + C2H3O2-(aq)</span><span>
</span><span>2NaOH(aq) + H2SO4 (aq) = Na2SO4 (s)+ 2H2O (aq)
</span>H2S (aq) + Ba(OH)2 (aq) = BaS (s)+ 2H2O (aq)
The answer is
a symmetrical molecule.
