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34kurt
3 years ago
7

A sample of an unknown gas occupies 5.51 L at 1.31 atm. What pressure would thisgas exert in a 0.520 L container if the temperat

ure is held constant?
Chemistry
1 answer:
weeeeeb [17]3 years ago
5 0

Answer:

P₂ = 13.9 atm (3 sig. figs.)

Explanation:

The pressure (P), Volume (V) relationship with Temperature (T) & mass (n) held constant is an inverse proportionality. That is Boyles Law ...

P ∝ 1/V => P = k/V => k = P·V

For two pressure-volume conditions, the proportionality constant (k) remains constant where k₁ = k₂ and P₁·V₁ = P₂·V₂ => P₂ = P₁·V₁/V₂

Given:

P₁ = 1.31 atm.

V₁ = 5.51 L

P₂ = ?

V₂ = 0.520 L

V₂ = (1.31 atm)(5.51L)/(0.520L) = 13.88096154 atm (calc. ans.) = 13.9 atm (3 sig. figs.)

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If 5.4 moles of Fe react with 4.7 moles of O2, what is the maximum amount of Fe2O3 (in moles) that can be produced? What is the
DIA [1.3K]

Answer:

2.7 moles of Fe₂O₃ is the maximum amount that can be produced. Iron is the limiting reactant.

Explanation:

The balanced reaction is:

4 Fe + 3 O₂ → 2 Fe₂O₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

  • Fe: 4 moles
  • O₂: 3 moles
  • Fe₂O3: 2 moles

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

You can use a simple rule of three as follows: if by stoichiometry 4 moles of Fe reacts with 3 moles of O₂, how much moles of Fe will be needed if 4.7 moles of O₂ react?

moles of Fe =\frac{4. moles of Fe*4.7 moles of O_{2}}{3 moles of O_{2} }

moles of O₂= 6.27

But 6.27 moles of Fe are not available, 5.4 moles are available. Since you have less moles than you need to react with 4.7 moles of O₂, iron Fe will be the limiting reagent.

So you can use a simple rule of three as follows: if by stoichiometry 4 moles of Fe produce 2 moles of Fe₂O₃, how many moles of Fe₂O₃ will be produced if 5.4 moles of Fe react?

moles of Fe_{2}O_{3}=\frac{5.4 moles of Fe*2 moles of Fe_{2} O_{3} }{4 moles of Fe}

moles of Fe₂O₃= 2.7 moles

Then:

<u><em>2.7 moles of Fe₂O₃ is the maximum amount that can be produced. Iron is the limiting reactant.</em></u>

4 0
3 years ago
Consider the combustion of octane:
just olya [345]

Answer:

15.9 g

Explanation:

(Take the atomic mass of C=12.0, H=1.0, O=16.0)

no. of moles = mass / molar mass

no. of moles of octane used = 11.2 / (12.0x8 + 1x18)

= 0.0982456 mol

Since oxygen is in excess and octane is the limiting reagent, the no. of moles of H2O depends on the no. of moles of octane used.

From the balanced equation, the mole ratio of octane : water = 2:18 = 1: 9,

so this means, one mole of octane produced 9 moles of water.

Using this ratio, we can deduce that (y is the no. of moles of water produced):

\frac{1}{9} =\frac{0.0982456}{y}

y = 0.0982456x9

y= 0.88421 mol

Since mass = no. of moles x molar mass,

mass of water produced = 0.88421  x (1.0x2+16.0)

=15.9 g

5 0
4 years ago
What mass of ammonium chloride (NH4Cl) would you use to prepare 85.0 mL of a 1.20M solution NH4Cl?
NeX [460]

Answer:

Form: M= mols solute/ liters solution

Ans: 5.46 NH4Cl

6 0
3 years ago
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ella [17]

Answer:

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Explanation:

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Answer:

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