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2 years ago

Acetylene gas (C2H2) burns completely in the presence of oxygen gas (O2) to yield carbon dioxide

Chemistry

1 answer:

Luda [366]2 years ago

8 0

A. The balanced equation for the reaction is

2C₂H₂ + 5O₂ —> 4CO₂ + 2H₂O

B. The volume of the oxygen gas required to burn 0.700 L of acetylene gas is 1.75 L

C. The volume of carbon dioxide gas produced is 1.4 L

D. The volume of water vapor produced is 0.7 L

<h3>A. Balanced equation </h3>

The balanced equation for the reaction between acetylene gas (C₂H₂) and oxygen gas (O₂) is given below

2C₂H₂ + 5O₂ —> 4CO₂ + 2H₂O

<h3>B. How to determine the volume of oxygen </h3>

2C₂H₂ + 5O₂ —> 4CO₂ + 2H₂O

Since the reaction occurred at standard temperature and pressure, we can thus say that:

From the balanced equation above,

2 L of C₂H₂ reacted with 5 L O₂.

Therefore,

0.7 L of C₂H₂ will react with = (0.7 × 5) / 2 = 1.75 L of O₂

Thus, 1.75 L of O₂ is needed for the reaction

<h3>C. How to determine the volume of carbon dioxide </h3>

2C₂H₂ + 5O₂ —> 4CO₂ + 2H₂O

From the balanced equation above,

2 L of C₂H₂ reacted to produce 4 L of CO₂

Therefore,

0.7 L of C₂H₂ will react to produce = (0.7 × 4) / 2 = 1.4 L of CO₂

Thus, 1.4 L of CO₂ were produced

<h3>D. How to determine the volume of water. </h3>

2C₂H₂ + 5O₂ —> 4CO₂ + 2H₂O

From the balanced equation above,

2 L of C₂H₂ reacted to produce 2 L of H₂O

Therefore,

0.7 L of C₂H₂ will also react to produce 0.7 L of H₂O

Thus, 0.7 L of H₂O was produced

Learn more about stoichiometry:

brainly.com/question/14735801

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A student is asked to standardize a solution of potassium hydroxide. He weighs out 1.08 g potassium hydrogen phthalate (KHC8H4O4

natka813 [3]

Answer:

A. 0.143 M

B. 0.0523 M

Explanation:

Let's consider the neutralization reaction between potassium hydroxide and potassium hydrogen phthalate (KHP).

KOH + KHC₈H₄O₄ → H₂O + K₂C₈H₄O₄

The molar mass of KHP is 204.22 g/mol. The moles corresponding to 1.08 g are:

1.08 g × (1 mol/204.22 g) = 5.28 × 10⁻³ mol

The molar ratio of KOH to KHC₈H₄O₄ is 1:1. The reacting moles of KOH are 5.28 × 10⁻³ moles.

5.28 × 10⁻³ moles of KOH occupy a volume of 36.8 mL. The molarity of the KOH solution is:

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Let's consider the neutralization of potassium hydroxide and perchloric acid.

KOH + HClO₄ → KClO₄ + H₂O

When the molar ratio of acid (A) to base (B) is 1:1, we can use the following expression.

$M_{A} \times V_{A} = M_{B} \times V_{B}\\M_{A} = \frac{M_{B} \times V_{B}}{V_{A}} \\M_{A} = \frac{0.143 M \times 10.1mL}{27.6mL}\\M_{A} =0.0523 M$

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