Use The Balanced Equation To Answer The Question

Use The Balanced Equation To Answer The Question2 C2h6g 5 O2g

Use the balanced chemical equations provided to determine the quantities of reactants and products involved in the reactions. Specifically, answer the following questions:

1. How many moles of O2 are required to completely react with 1.8 moles of C2H6?

2. How many moles of H2O are produced from 0.10 moles of C2H6?

3. How many moles of C2H6 are necessary to produce 9.2 moles of CO?

Additionally, using the combustion reaction of acetylene (C2H2), answer:

a. How many moles of O2 are needed to fully react with 7.40 mol of C2H2?

b. How many moles of C2H2 are needed to produce 0.60 mol of CO2?

c. How many moles of CO2 are generated from 9.0 mol of C2H2?

d. How many moles of H2O are produced from 0.13 mol of C2H2?

Ensure to use the stoichiometric ratios from the balanced equations to calculate each answer accurately.

Sample Paper For Above instruction

Use The Balanced Equation To Answer The Question2 C2h6g 5 O2g

Use The Balanced Equation To Answer The Question2 C2h6g 5 O2g

Combustion reactions are fundamental in chemistry, illustrating how hydrocarbons react with oxygen to produce carbon monoxide, dioxide, water, and other products. The balanced equations guide the quantitative analysis necessary for predicting the amounts of reactants needed and products formed. Here, we analyze the combustion of ethane (C2H6) and acetylene (C2H2) to address several questions involving molar ratios.

Ethane Combustion Analysis

The balanced chemical equation for ethane combustion is:

2 C2H6 (g) + 5 O2 (g) → 4 CO (g) + 6 H2O (g)

From this balanced equation, the molar ratios are evident. Each 2 moles of ethane react with 5 moles of oxygen to produce 4 moles of carbon monoxide and 6 moles of water vapor. Using these ratios, calculations are as follows:

1. Moles of O2 needed to react with 1.8 moles of C2H6

The ratio from the equation is 2 mol C2H6 : 5 mol O2. Therefore, for 1.8 mol C2H6:

O2 needed = (5 mol O2 / 2 mol C2H6) × 1.8 mol C2H6 = 4.5 mol O2

2. Moles of H2O produced from 0.10 mol C2H6

The ratio is 2 mol C2H6 : 6 mol H2O. So, for 0.10 mol C2H6:

H2O produced = (6 mol H2O / 2 mol C2H6) × 0.10 mol C2H6 = 0.30 mol H2O

3. Moles of C2H6 needed to produce 9.2 mol of CO

The ratio is 2 mol C2H6 : 4 mol CO. Thus, the amount of C2H6 required:

C2H6 needed = (2 mol C2H6 / 4 mol CO) × 9.2 mol CO = 4.6 mol C2H6

Analysis of Acetylene (C2H2) Combustion

The balanced chemical equation is:

2 H-C≡C-H + 5 O2 → 4 CO2 + 2 H2O

From this, the molar ratios are extracted for various calculations:

a. O2 needed to react with 7.40 mol C2H2

Ratio: 2 mol C2H2 : 5 mol O2

O2 needed = (5 mol O2 / 2 mol C2H2) × 7.40 mol C2H2 = 18.5 mol O2

b. C2H2 needed to produce 0.60 mol CO2

Ratio: 2 mol C2H2 : 4 mol CO2

C2H2 needed = (2 mol C2H2 / 4 mol CO2) × 0.60 mol CO2 = 0.30 mol C2H2

c. Moles of CO2 from 9.0 mol C2H2

Ratio: 2 mol C2H2 : 4 mol CO2

CO2 formed = (4 mol CO2 / 2 mol C2H2) × 9.0 mol C2H2 = 18 mol CO2

d. Moles of H2O from 0.13 mol C2H2

Ratio: 2 mol C2H2 : 2 mol H2O

H2O formed = (2 mol H2O / 2 mol C2H2) × 0.13 mol C2H2 = 0.13 mol H2O

Conclusion

The calculations demonstrate how balanced chemical equations serve as vital tools for quantitative predictions in combustion chemistry. Such analyses are essential for industrial applications, environmental assessments, and theoretical studies involving hydrocarbons.

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