Answer:
1) There were 7.65 grams of heptane burned
2) There reacted 38.94 grams of HCl
Explanation:
<em>1) The combustion of heptane</em>
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
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<em>2) The reaction of limestone with hydrochloric acid</em>
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
