Answer:
The answers to the question are
(a) 13.24 g
(b) (O₂)
(c) 4.8252 g
(2) 0.662 M/L
Explanation:
(a) To solve the question we write the equation as follows
C + O₂ → CO₂
That is one mole of graphite reacts with one mole of oxygen to form one mole of carbon dioxide
number of moles of graphite = 8.44/12 = 0.703 moles
number of moles of oxygen = 9.63/32 = 0.3009
However since one mole of graphite reacts with one mole of oxygen to form one mole of carbon dioxide, therefore, 0.3009 moles of oxygen will react with 0.3009 moles of carbon to fore 0.3009 moles of CO₂
The maximum mass of carbon dioxide that can be formed = mass = moles × molar mass
= 0.3009×44 = 13.24 g
(b) The formula for the limiting reagent (O₂)
Finding the limting reagent is by checking the mole balance of the reactants available to the moles specified in th stoichiometry of the reaction and selecting the reagent with the list number of moles
(c) The mass of excess reagent = 0.703 moles - 0.3009 moles = 0.4021 moles
mass of excess reagent = 0.4021 × 12 = 4.8252 g
(2) The molarity is given by number of moles per liter of solution, therefore
molar mass of mgnesium iodide = 278.1139 g/mol, number of moles of magnesium iodide in 23 g = 23g/ 278.1139 g/mol= 8.3 × 10⁻² M
Therefore the moles in 125 mL = (8.3 × 10⁻² M)/(125 mL) = (8.3 × 10⁻² M)/(0.125 L) = 0.662 M/L
