Answer:
−6063.54 J of heat is removed from the system during this process.
Explanation:
Given that :
in an isothermal process
number of moles of an ideal gas = 1.59 moles
which is compressed to one- fifth of its initial volume
Let say the initial volume = x
the final volume will be = x/5
Temperature = 285 K
The objective is to determine what quantity of heat is added to, or removed from, the system during this process.
In an isothermal process;
q = w
where ;
w = nRT In (V₂/V₁)
w = 1.59 × 8.314 × 285 × In (x/5 /x)
w = 1.59 × 8.314 × 285 × In (1/5)
w = −6063.54 J
Thus; −6063.54 J of heat is removed from the system during this process.
Answer:
KO is the limiting reactant.
0.11 mol O₂ will be produced.
Explanation:
4 KO₂ + 2 H₂O ⇒ 4 KOH + 3 O₂
Find the limiting reagent by dividing the moles of the reactant by the coefficient in the equation.
(0.15 mol KO₂)/4 = 0.0375
(0.10 mol H₂O)/2 = 0.05
KO₂ is the limiting reagent.
The amount of product produced depends on the limiting reagent. To find how much is produced, take moles of limiting reagent and multiply it by the ratio of reagent to product. You can find the ratio by looking at the equation. For every 4 moles of KO₂, 3 moles of O₂ are produced.
0.15 mol KO₂ (3 mol O₂)/(4 mol KO₂) = 0.1125 mol O₂
0.11 mol O are produced.
The reaction is,
H2S + I2 --------------> 2 HI +S
Molar weight of H2S = 34 g per mol
Molar weight of HI =128 g per mol
Molar weight of I2 =254 g per mol
Moles of H2S in 49.2 g = 49.2 /34 mol = 1.447 mol
So according to stoichiometry of the reaction, number of I2 mols needed
= 1.447 mol
The mass of I2 needed = 1.447 mol x 254 g
Moles= molarity x liter
=0.2500 M x 0.500 L
= 0.1250 mol
mass C2H2O4 . 2H2O = 0.1250 mol x (126.068 g / 1 mol)
=15.76 g
