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BlackZzzverrR [31]

3 years ago

7

A gas mixture of Ne and Ar has a total pressure of 4.00 atm and contains 16.0 mol of gas. If the partial pressure of Ne is 2.75

atm, how many moles of Ar are in the mixture?
A) 11.0
B) 5.00
C) 6.75
D) 9.25
E) 12.0

1 answer:

Stolb23 [73]3 years ago

4 0

Answer:

5 moles of Argon is present in the mixture.

Explanation:

Total pressure of the gaseous mixture = 4 atm

Total number of moles = 16

Partial pressure of Ne = 2.75 atm

By Dalton's law of partial pressure, the total pressure of gaseous mixture is the sum of partial pressures of individual gases which are non-reactive.

Hence:

$P_{total}=P_{Ar}+P_{Ne}\\4=P_{Ar}+2.75\\P_{Ar}=1.25\ atm$

Also :

Partial pressure = mole fraction*total pressure

$P_{Ar}=X_{Ar}P_{total}$

$X_{Ar}=\frac{1.25}{4}=0.3125$

$\frac{n_{Ar}}{n_{total}}=0.3125\\n_{Ar}=5$

∴Number of moles of Argon = 5

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Consider the reaction Mg₂Si(s) + 4 H₂O(ℓ) → 2 Mg(OH)₂(aq) + SiH₄(g). How many grams of silane gas (SiH₄) are formed if 25.0 g of

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Answer:

10.60 grams of silane gas are formed.

Explanation:

From the reaction:

Mg₂Si(s) + 4H₂O(l) → 2Mg(OH)₂(aq) + SiH₄(g)

We know that the limiting reactant is Mg₂Si, so to find the mass of SiH₄ formed we need to calculate the number of moles of Mg₂Si:

$\eta_{Mg_{2}Si} = \frac{m_{Mg_{2}Si}}{M_{Mg_{2}Si}}$

Where:

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$\eta_{Mg_{2}Si} = \frac{m_{Mg_{2}Si}}{M_{Mg_{2}Si}} = \frac{25.0 g}{76.69 g/mol} = 0.33 moles$

Now, the stoichiometric relation between Mg₂Si and SiH₄ is 1:1 so:

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According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The chemical reaction for the formation reaction of $AlCl_3$ is:

$2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3 (s)$ $\Delta H^o_{formation}=?$

The intermediate balanced chemical reaction are:

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