Chemistry College

The balanced reaction equation for combustion of heptane, C 7 H 16 , is C 7 H 16 + 11 O 2 ⟶ 7 CO 2 + 8 H 2 O If the reaction produced 23.5 g CO 2 , how many grams of heptane were burned? mass: g C 7 H 16 The reaction of limestone with hydrochloric acid is CaCO 3 + 2 HCl ⟶ CaCl 2 + CO 2 + H 2 O If the reaction produced 23.5 g CO 2 , how many grams of HCl reacted? mass: g HCl

Answers

Answer 1

Answer:

1) There were 7.65 grams of heptane burned

2) There reacted 38.94 grams of HCl

Explanation:

1) The combustion of heptane

Step 1: Data given

Mass of CO2 = 23.5 grams

Molar mass of CO2 = 44.01 g/mol

Step 2: The balanced equation:

C7H16  + 11O2 ⟶ 7CO2 + 8H2O

Step 3: Calculate moles of CO2

Moles CO2 = mass CO2 / molar mass CO2

Moles CO2 = 23.5 grams / 44.01 g/mol

Moles CO2 = 0.534 moles

Step 4: Calculate moles heptane

For 1 mole of Heptane , we need 11 moles of O2 to produce 7 moles of CO2 and 8 moles of H2O

For 0.534 moles of CO2 we have 0.534/7 = 0.0763 moles of  heptane

Step 5: Calculate mass of heptane

Mass of heptane = moles heptane * molar mass heptane

Mass heptane = 0.0763 moles * 100.21 g/mol

Mass heptane = 7.65 grams

2) The reaction of limestone with hydrochloric acid

Step 1: Data given

Mass of CO2 = 23.5 grams

Step 2: The balanced equation:

CaCO3 + 2HCl ⟶ CaCl2 + CO2 + H2O

Step 3: Calculate moles of CO2

Moles CO2 = mass CO2 / molar mass CO2

Moles CO2 = 23.5 grams / 44.01 g/mol

Moles CO2 = 0.534 moles

Step 4: Calculate moles of HCl

For 1 mol of CaCO3 we need 2 moles of HCl to produce 1 mol of CaCl2, 1 mol of CO2 and 1 mol of H2O

For 0.534 moles of CO2 we have 2*0.534 = 1.068 moles of HCl

Step 5: Calculate mass of HCl

Mass HCl = moles HCl * molar mass HCl

Mass HCl = 1.068 moles * 36.46 g/mol

Mass HCl = 38.94 grams

Answer 2

To produce 23.5 g of CO₂, approximately 7.63 g of heptane were burned, and 38.94 g of HCl reacted. By using the molar masses and stoichiometric relationships from the balanced reactions, we can find the required masses. Moles of substances involved were calculated using the molar masses and balanced equations.

Determining Grams of Heptane and HCl Reacted

First, let's find out how many grams of heptane (C₇H₁₆) were burned to produce 23.5 g of CO₂ in the given balanced chemical reaction:

Combustion of Heptane:

Balanced equation: C₇H₁₆ + 11 O₂ ⟶ 7 CO₂ + 8 H₂O

Molar mass of CO₂: 12.01 (C) + 2 × 16.00 (O) = 44.01 g/molMoles of CO₂ produced: 23.5 g ÷ 44.01 g/mol ≈ 0.534 moles CO₂From the balanced equation, 7 moles of CO₂ are produced per 1 mole of heptane. Thus, moles of heptane burned: 0.534 moles CO₂ ÷ 7 ≈ 0.076 moles heptane.Molar mass of heptane (C₇H₁₆): 7 × 12.01 (C) + 16 × 1.01 (H) = 100.23 g/molGrams of heptane burned: 0.076 moles × 100.23 g/mol ≈ 7.63 g.

Next, let's determine how many grams of HCl reacted to produce 23.5 g of CO₂ in the reaction between limestone and hydrochloric acid:

Reaction of Limestone with Hydrochloric Acid:

Balanced equation: CaCO₃ + 2 HCl ⟶ CaCl₂ + CO₂ + H₂O

Moles of CO₂ produced: 23.5 g ÷ 44.01 g/mol ≈ 0.534 moles CO₂From the balanced equation, 1 mole of CO₂ is produced per 2 moles of HCl. Thus, moles of HCl reacted: 0.534 moles CO₂ × 2 ≈ 1.068 moles HCl.Molar mass of HCl: 1.01 (H) + 35.45 (Cl) = 36.46 g/molGrams of HCl reacted: 1.068 moles × 36.46 g/mol ≈ 38.94 g.

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