The empirical formula of metal iodide : CoI₃(Cobalt(III) Iodide)
<h3>Further explanation</h3>
13.02 g sample of Cobalt , then mol Co(MW=58.933 g/mol) :
Mass of metal iodide formed : 97.12 g, so mass of Iodine :
Then mol iodine (MW=126.9045 g/mol) :
mol ratio of Cobalt and Iodine in the compound :
C. Rutherford would be the answer
Answer:
C. NaHCO3 → Na2CO3 + H2O + CO2
Explanation:
» For a balanced equation, the number of atoms of reactants and products must be equal.
» In equation C, the reactant side has one sodium atom while ptoduct side has 2 sodium atoms.
» The balanced equations mus be;
<u>Given:</u>
Moles of Al = 0.4
Moles of O2 = 0.4
<u>To determine:</u>
Moles of Al2O3 produced
<u>Explanation:</u>
4Al + 3O2 → 2Al2O3
Based on the reaction stoichiometry:
4 moles of Al produces 2 moles of Al2O3
Therefore, 0.4 moles of Al will produce:
0.4 moles Al * 2 moles Al2O3/4 moles Al = 0.2 moles Al2O3
Similarly;
3 moles O2 produces 2 moles Al2O3
0.4 moles of O2 will yield: 0.4 *2/3 = 0.267 moles
Thus Al will be the limiting reactant.
Ans: Maximum moles of Al2O3 = 0.2 moles
B. I’m sorry if I’m wrong ! Wasn’t sure about this question .
