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chubhunter [2.5K]
3 years ago
13

A scientist needs to collect a 0.050 mole sample of helium, but needs to know how large his helium container should be. What vol

ume is needed to store the sample of helium gas at 202.6kPa and 400K? A. 657 milliliters B. 760 milliliters C. 820 milliliters D. 1,025 milliliters
Chemistry
1 answer:
Mumz [18]3 years ago
8 0

We can use the ideal gas equation:

PV = nRT

P = 202.6kPa = 202600 Pa (You have to multiply by 1000)

n = 0.050 mole

R = 0.082 atm*l/(K*mol)

T = 400K

We will have to convert from Pa to atm or viceversa.

101325 Pa________1 atm

202600 Pa________x = 2.00 atm

2atm*V = 0.050 mole*0.082 atm*l/(K*mol)* 400K

V = 0.050 mole*0.082 atm*l/(K*mol)* 400K/2atm = 0.82 liters = 820 mililiters



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NH3 is the limiting reactant

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<u>Step 2:</u> The balanced equation:

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<u>Step 3:</u> Calculate moles of NH3

Moles NH3 = Mass NH3 / Molar Mass NH3

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<u>Step 4:</u> Calculate moles of O2

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<u>Step 5: </u>Calculate the limiting reactant

For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

NH3 is the limiting reactant. It will completely be consumed ( 0.126 moles).

O2 is in excess, there will be 3/4 * 0.126 = 0.0945 moles consumed

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For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

For 4 moles NH3 , we'll have 2 moles of N2 produced

For 0.126 moles NH3 consumed, we'll have 0.063 moles of N2 produced.

<u>Step 7</u>: Calculate volume of N2 produced

p*V = n*R*T

⇒ with p = the pressure of the gas = 1.00 atm

⇒ with V = the volume = TO BE DETERMINED

⇒ with n = the number of moles N2 = 0.063 moles

⇒ with R = the gasconstant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 295

V = (nRT)/p

V = (0.063*0.08206*295)/1

V = 1.525 L = theoretical yield

<u>Step 8:</u> Calculate the % yield

% yield = actual yield / theoretical yield

% yield = (0.550 L / 1.525 L)*100%

% yield = 36.1 %

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