<span>the balanced equation for the combustion of pentane is as follows:
C5H12 + 8O2 --> 5CO2 + 6H2O
excess pentane is allowed to react with 4 moles of oxygen. this means that pentane is the excess reactant and O</span>₂<span> is the limiting reactant. Limiting reactant is the reagent that is fully used up in the reaction and amount of product formed depends on the amount of limiting reactant present. Stoichiometry of O</span>₂<span> to H</span>₂<span>O is 8: 6. If 8 mol of O</span>₂<span> forms 6 mol of H</span>₂<span>O then 4 mol of O</span>₂<span> forms - 6/8 x 4 = 3 mol of H</span>₂<span>O therefore 3 mol of H</span>₂<span>O is formed</span>
Answer: 3 <span>moles of water would be produced in present case. </span> Reason: Reaction involved in present case is: <span> C5H12 + 8O2 </span>→<span> 5CO2 + 6H2O
In above reaction, 1 mole of C5H12 reacts with 8 moles of oxygen to give 6 moles of water.
Thus, 4 moles of oxygen will react with 0.5 mole of C5H12, to generate 3 moles of H2O.</span>
Flerovium at its ground state is solid. It has electron configuration of [Rn]5f¹⁴6d¹⁰7s²7p². The expected number of valence electrons in a flerovium atom is 2. A ground state is the most stable state of an atom at satndard temperature and pressure.
