Answer:
6.4×10¯³ g of O₂.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
CH₄ + 2O₂ —> CO₂ + 2H₂O
Next, we shall determine the masses of CH₄ and O₂ that reacted from the balanced equation. This can be obtained as follow:
Molar mass of CH₄ = 12 + (4×1)
= 12 + 4
= 16 g/mol
Mass of CH₄ from the balanced equation = 1 × 16 = 16 g
Molar mass of O₂ = 2 × 16 = 32 g/mol
Mass of O₂ from the balanced equation = 2 × 32 = 64 g
SUMMARY:
From the balanced equation above,
16 g of CH₄ reacted with 64 g of O₂.
Finally, we shall determine the mass of O₂ needed to react with 1.6×10¯³ g of CH₄. This can be obtained as illustrated below:
From the balanced equation above,
16 g of CH₄ reacted with 64 g of O₂.
Therefore, 1.6×10¯³ g of CH₄ will react with = (1.6×10¯³ × 64) / 16 = 6.4×10¯³ g of O₂
Thus, 6.4×10¯³ g of O₂ is needed for the reaction.
The description of an ideal gas is as follows:
- They have a small mass
- They have a small volume
- They have no intermolecular forces
<h3>WHAT IS AN IDEAL GAS?</h3>
An ideal gas is a theoretical gas composed of many randomly moving particles that do not engage in any inter-particle interactions.
An ideal gas obeys all the gas laws and possess the following characteristics:
- They have a small mass
- They have a small volume
- They have no intermolecular forces
Learn more about gas laws at: brainly.com/question/1437490
Answer:
0.26 mol
Explanation:
using general gas equation
PV=nRT
V=4.1litre= 4.1 dm³
P= 1.78 atm
R= 0.0821
PUT VALUES
