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aleksley [76]
3 years ago
10

Can someone help me please? I’d appreciate it! Also please do not send me a link or delete your answer! Thanks

Chemistry
1 answer:
chubhunter [2.5K]3 years ago
5 0

Answer:

14.53ML

Explanation:

V1=218

V2=?

P2=15p1

USING BOYLE'S LAW

P1V1=P2V2

V2=P1V1/P2=P1(218ML)/15P1

=14.53ML

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<em>Full question: Assume no volume change.  If you formed 0.0910 atm of gas, what is the percent yield?</em>

<em />

The reaction that is occurring is:

Ni + 3HBr → NiBr₃ + 3/2H₂(g)

First, we will determine moles of Ni and HBr to determine limiting reactant and theoretical yield

Using ideal gas law, we can determine the moles of hydrogen formed. Thus, we can find percent yield:

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As there are just 0.01738 moles of Ni, the Ni is limiting reactant. Assuming a theoretical yield, moles of H₂ produced are:

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Now, moles of H₂ produced are:

PV = nRT

PV/RT = n

<em>Where P is pressure (0.0910atm)</em>

<em>V is volume (2.50L)</em>

<em>R is gas constant (0.082atmL/molK)</em>

<em>T is absolute temperature in Kelvin (30°C + 273.15 = 303.15K)</em>

<em>And n are moles</em>

PV/RT = n

0.0910atm*2.50L/0.082atmL/molK*303.15K = n

0.00915 moles = n

<em />

And percent yield (Produced moles / Theoretical moles * 100) is:

0.00915 moles / 0.02607moles =

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