Based on the data provided;
- number of moles of helium gas is 1.25 moles
- pressure at peak temperature is 259.3 kPa
- internal pressure is above 256 kPa, therefore, the balloon will burst.
- pressure should be reduced to a value less than 256 kPa by reducing the temperature
<h3>What is the ideal has equation?</h3>
The ideal gas equation relatesthe pressure, volume, moles and temperature of a gas.
The moles of helium gas is calculated using the Ideal gas equation:
n is the number of moles of gas
R is molar gas constant = 8.314 L⋅kPa/Kmol
P is pressure = 239 kPa
T is temperature = 21°C = 294 K
V is volume = 12.8 L
Therefore;
n = PV/RT
n = 239 × 12.8 / 8.314 × 294
n = 1.25 moles
The number of moles of helium gas is 1.25 moles
At peak temperature, T = 46°C = 319 K
Using P1/T1 = P2/T2
P2 = P1T2/T1
P2 = 239 × 319/294
P2 = 259.3 kPa
The pressure at peak temperature is 259.3 kPa
At 42°C, T = 315 K
Using P1/T1 = P2/T2
P2 = P1T2/T1
P2 = 239 × 315/294
P2 = 256.07 kPa
Since the internal pressure is above 256 kPa, the balloon will burst.
The pressure should be reduced to a value less than 256 kPa by reducing the temperature.
Learn more about gas ideal gas equation at: brainly.com/question/12873752
First, let's compute the number of moles in the system assuming ideal gas behavior.
PV = nRT
(663 mmHg)(1atm/760 mmHg)(60 L) = n(0.0821 L-atm/mol-K)(20+273 K)
Solving for n,
n = 2.176 moles
At standard conditions, the standard molar volume is 22.4 L/mol. Thus,
Standard volume = 22.4 L/mol * 2.176 mol =<em> 48.74 L</em>
<span>atomic weights: Al = 26.98, Cl = 35.45
In this reaction; 2Al = 53.96 and 3Cl2 = 212.7
Ratio of Al:Cl = 53.96/212.7 = 0.2537 that is approximately four times the mass Cl is needed.
Step 2:
(a) Ratio of Al:Cl = 2.70/4.05 = 0.6667
since the ratio is greater than 0.2537 the divisor which is Cl is not big enough to give a smaller ratio equal to 0.2537.
so Cl is limiting
(b)since Cl is the limiting reactant 4.05g will be used to determine the mass of AlCl3 that can be produced.
From Step 1:
212.7g of Cl will produce 266.66g AlCl3
212.7g = 266.66g
4.05g = x
x = 5.08g of AlCl3 can be produced
(c)
Al:Cl = 0.2537
Al:Cl = Al:4.05 = 0.2537
mass of Al used in reaction = 4.05 x 0.2537 = 1.027g
Excess reactant = 2.70 - 1.027 = 1.67g
King Leo · 9 years ago</span>
