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Alina [70]
3 years ago
10

How many moles of a gas would occupy 11.4 L at 273K and 2.00 atm?

Chemistry
1 answer:
bagirrra123 [75]3 years ago
3 0

Answer:

1.02mol

Explanation:

Using the general gas equation below;

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (K)

According to the information provided in this question,

P = 2.0 atm

V = 11.4L

T = 273K

n = ?

Using PV = nRT

n = PV/RT

n = 2 × 11.4/ 0.0821 × 273

n = 22.8/22.41

n = 1.017

n = 1.02mol

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In this stoichiometry problem, determine the limiting reactant.
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Answer:

Limiting reactant is NiSO₄

Explanation:

The reaction of aluminum metal with aqueous nickel(II)  sulfate to produce aqueous aluminum sulfate and nickel is:

2 Al(s) + 3 NiSO₄ → Al₂(SO₄)₃ + 3 Ni  

<em>That means 2 moles of Al react with 3 moles of nickel sulfate.</em>

<em />

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According to stoichiometry :

2 moles of NaClO_3 give = 2 moles of NaCl

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