Calculate the amount of carbon dioxide that could be produced when

The balanced reaction of combustion of carbon can be written as: (i) As per the balanced equation, 1 mole of carbon burns in1 mole of dioxygen (air) to produce1 mole of carbon dioxide. (ii) According to the question, only 16 g of dioxygen is available. Hence, it will react with 0.5 mole of carbon to give 22 g of carbon dioxide. Hence, it is a limiting reactant. (iii) According to the question, only 16 g of dioxygen is available. It is a limiting reactant. Thus, 16 g of dioxygen can combine with only 0.5 mole of carbon to give 22 g of carbon dioxide.

Liquid octane CH3CH26CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 8.0 g of octane is mixed with 8.96 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits. This is not much different from your previous questions. Here goes one more time. BTW, octane should be CH 3 (CH 2 ) 6 CH 3 which is also C 8 H 18 Write the correctly balanced equation for the reaction: 2C 8H 18 + 25 O 2 ==> 16CO 2 + 18H 2O ... balanced equation for combustion of octane Since we are given the amounts of BOTH reactants, we must find which one is limiting. moles octane = 8.0 g C 8H 18 x 1 mol/114 g = 0.070 moles C 8H 18 moles O 2 = 8.96 g O 2 x 1 mol/32 g = 0.28 moles O 2 --> LIMITING REACTANT max mass CO 2 = 0.28 mol O 2 x 16 mol CO 2 / 25 mol O 2 x 44 g CO 2 / mol CO 2 = 7.9 g CO 2 (2 sig. figs.)

Balanced Equation #"3NaHCO"_3+"H"_3"C"_6"H"_5"O"_7"# #rarr# #"Na"_3"C"_6"H"_5"O"_7+"3CO"_2+"3H"_2"O"# We will need the molar masses of #"NaHCO"_3"# and #"CO"_2"#, and their mole ratio as well. Molar Masses #"NaHCO"_3":# #"84.006609 g/mol"# #"CO"_2":# #"44.0095 g/mol"# Mole Ratio From the balanced equation, #"NaHCO"_3"# and #"CO"_2"# is #"3 mol NaHCO"_3:# #"3 mol CO"_2"#. Stoichiometric Equation Determine the moles of #"NaHCO"_3"# by dividing the given mass of #"NaHCO"_3"# by its molar mass, then determine moles of #"CO"_2"# by multiplying times the mole ratio with #"CO"_2"# in the numerator, then multiply times the molar mass of #"CO"_2"#. #2.50cancel"g NaHCO"_3xx(1cancel"mol NaHCO"_3)/(84.006609cancel"g NaHCO"_3)xx(cancel3^1cancel"mol CO"_2)/(cancel3^1cancel"mol NaHCO"_3)xx(44.0095"g CO"_2)/(1cancel"mol CO"_2)="1.31 g CO"_2"# rounded to three

Answer: The balanced reaction of combustion of carbon can be written as: C + O2----------------> CO2 1 1 1 (i) As per the balanced equation, 1 mole of carbon burns in 1 mole of dioxygen (air) to produce 1 mole of carbon dioxide. (ii) In this part mass of two reactants are given so we need to find the limiting reagent first Atomic mass of C = 12 amu Molecular mass of O 2= 2 × 16.00 = 32 amu According to Stoichiometry of the reaction 12 g C will react with 32 g of O 2 But we have only 16 g O 2 Given mass of O 2(16 g) < required mass of O 2(32g) Hence O 2is limiting reagent and product is always calculated with the help of mass of limiting reagent Molecular mass of CO 2= 12 + 2× 16 = 44 amu From the reaction 32 g of O 2produce = 44 g CO 2 1 g of O 2Produce = 44/32 g CO2 16 g O 2will produce = 44 × 16 / 32 = 22 g CO 2 Hence, 22 g ofcarbon dioxide is produced when1 mole of carbon is burnt in 16 g of dioxygen. (iii) According to the question, only 16 g of dioxygen is available. It is a limiting reactant. In this part mass of two reactants are given so we need to find the limiting reagent first Atomic mass of C = 12 amu Molecular mass of O 2= 2 × 16.00 = 32 amu 2 mol of C = 2 x 12 = 24 g According to Stoichiometry of the reaction 12 g C will react with = 32 g of O 2 1 g of C will react with = 32/ 12 g O 2 24 g of C will react with = 24 × 32 / 12 = 64 g O 2 But we have only 16 g O 2 Given mass of O 2(16 g) < required mass of O 2(64g) Hence O 2is limiting reagent and product is alw...