Question

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

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