Answer:
71.6 g of Ru(CO)₅ is the maximum mass that can be formed.
The limiting reactant is Ag
Explanation:
The reaction is:
RuI₃ (s) + 5CO (g) + 3Ag (s) → Ru(CO)₅ (s) + 3AgI (s)
Firstly we determine the moles of each reactant:
169 g . 1mol /481.77g = 0.351 moles of RuI₃
58g . 1mol /28g = 2.07 moles of CO
96.2g . 1mol/ 107.87g = 0.892 moles
Certainly, the excess reactant is CO, therefore, the limiting would be Ag or RuI₃.
3 moles of Ag react to 1 mol of RuI₃
Then 0.892 moles of Ag may react to (0.892 . 1) /3 = 0.297 moles
We have 0.351 moles of iodide and we need 0.297 moles, so this is an excess. In conclussion, Silver (Ag) is the limiting.
1 mol of RuI₃ react to 3 moles of Ag
Then, 0.351 moles of RuI₃ may react to (0.351 . 3) /1 = 1.053 moles
It's ok, because we do not have enough Ag. We only have 0.892 moles and we need 1.053.
5 moles of CO react to 3 moles of Ag
Then, 2.07 moles of CO may react to (2.07 . 3) /5 = 1.242 moles of Ag.
This calculate confirms the theory.
Now, we determine the maximum mass of Ru(CO)₅
3 moles of of Ag can produce 1 mol of Ru(CO)₅
Then 0.892 moles may produce (0.892 . 1) /3 = 0.297 moles
We convert moles to mass → 0.297 mol . 241.07g /mol = 71.6 g
