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Basile [38]
3 years ago
9

What is the volume of 1.9 moles of chlorine gas (Cl2) at standard temperature and pressure (STP)?

Chemistry
2 answers:
masya89 [10]3 years ago
5 0
Considering ideal gas behavior, the volume of 1 mol of gas at STP is 22.4 L; then the volume occupied by 1.9 moles is 1.9mol*22.4L/mol = 42. 6 L.

Answer: 43 L
nexus9112 [7]3 years ago
4 0

Answer : The volume of chlorine gas is, 42.56 L

Solution : Given,

Moles of chlorine gas = 1.9 moles

As we know that,

At STP, 1 mole contains 22.4 L volume of gas

As, 1 mole of chlorine gas contains 22.4 L volume of chlorine gas

So, 1.9 moles of chlorine gas contains 22.4\times 1.9=42.56L volume of chlorine gas

Therefore, the volume of chlorine gas is, 42.56 L

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If 175mL of oxygen is produced at STP, how many grams of hydrogen peroxide, H2O2
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Answer:

0.53g

Explanation:

We'll begin by converting 175mL to L. This is illustrated below:

1000mL = 1L

Therefore 175mL = 175/1000 = 0.175L

Next, we shall calculate the number of mole of O2 that occupy 0.175L. This is illustrated below:

1 mole of O2 occupy 22.4L at stp.

Therefore, Xmol of O2 will occupy 0.175L i.e

Xmol of O2 = 0.175/22.4

Xmol of O2 = 7.81×10¯³ mole

Therefore, 7.81×10¯³ mole of O2 occupy 175mL.

Next, we shall determine the number of mole of H2O2 that decomposed to produce 7.81×10¯³ mole of O2. This is illustrated below:

2H2O2 —> 2H2O + O2

From the balanced equation above,

2 moles of H2O2 decomposed to produce 1 mole of O2.

Therefore, Xmol of H2O2 will decompose to produce 7.81×10¯³ mole of O2 i.e

Xmol of H2O2 = 2 x 7.81×10¯³

Xmol of H2O2 = 1.562×10¯² mole

Therefore, 1.562×10¯² mole of H2O2 decomposed in the reaction.

Finally, we shall convert 1.562×10¯² mole of H2O2 to grams. This is illustrated below:

Molar mass of H2O2 = (2x1) + (16x2) = 34g/mol

Mole of H2O2 = 1.562×10¯² mole

Mass of H2O2 =..?

Mole = mass /Molar mass

1.562×10¯² = mass /34

Cross multiply

Mass of H2O2 = 1.562×10¯² x 34

Mass of H2O2 = 0.53g

Therefore, 0.53g of Hydrogen peroxide, H2O2 were decomposition in the reaction.

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