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Pavel [41]
3 years ago
8

If a container is filled with 13.3 L of gas at 1.41 atm. The container is held at constant temperature throughout the experiment

if the pressure changes to 1.02 atm what will the final volume be
Chemistry
1 answer:
laila [671]3 years ago
3 0

Answer:

The final volume is 18.38 L.

Explanation:

It is given that,

A container is filled with 13.3 L of gas at 1.41 atm. The container is held at constant temperature throughout the experiment if the pressure changes to 1.02 atm.

It is based on Boyle's law. According to this law, at constant temperature,

PV=K

K is constant

P_1V_1=P_2V_2

V_2=\dfrac{P_1V_1}{P_2}\\\\V_2=\dfrac{13.3\times 1.41}{1.02}\\\\V_2=18.38\ L

So, the final volume is 18.38 L.

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Sodium hydroxide reacts with aluminum and water to produce hydrogen gas: 2 Al(s) + 2 NaOH(aq) + 6 H2O(l) → 2 NaAl(OH)4(aq) + 3 H
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Answer:

The mass of hydrogen gas formed is 0.205 grams

Explanation:

<u>Step 1:</u> Data given

Mass of 1.83 grams of Al

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Molar mass of Al = 26.98 g/mol

Molar mass of NaOH = 40 g/mol

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

2 Al(s) + 2 NaOH(aq) + 6 H2O(l) → 2 NaAl(OH)4(aq) + 3 H2(g)

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

Moles Al = mass Al / Molar mass Al

Moles Al = 1.83 grams / 26.98 g/mol

Moles Al = 0.0678 moles

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

Moles NaOH = 4.30 grams / 40 g/mol

Moles NaOH = 0.1075 moles

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

For 2 moles of Al, we need 2 moles of NaOH

Aluminium is the limiting reactant. It will completely be consumed ( 0.0678 moles)

NaOH is in excess. There will react 0.0678 moles

There will remain 0.1075 - 0.0678 = 0.0397 moles

<u>Step 6</u>: Calculate moles of hydrogen

For 2 moles of Al, we need 2 moles of NaOH, to produce 3 moles of hydrogen

For 0.0678 moles of Al, there is produced 0.0678 *3/2 = 0.1017 moles of H2

<u>Step 7</u>: Calculate mass of H2

Mass of H2 = Moles H2 * Molar mass of H2

Mass of H2 = 0.1017 moles * 2.02 g/mol

Mass of H2 = 0.205 grams

The mass of hydrogen gas formed is 0.205 grams

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