15.0 moles of silver chloride are produced.
Answer:
0.190L of hydrogen may be produced by the reaction.
Explanation:
Our reaction is:
3Mg + 2H₃PO₄ → Mg₃(PO₄)₂ + 3H₂
We need to determine the limting reactant. Let's find out the moles of each:
5.159×10²¹ atoms . 1 mol / 6.02×10²³ atoms = 0.00857 moles of Mg
55.23 g . 1 mol / 97.97 g = 0.563 moles of acid
2 moles of acid react to 3 moles of Mg
0.563 moles of acid may react to: (0.563 . 3) /2 = 0.8445 moles of Mg
Definetely the limting reactant is Mg.
As ratio is 3:3, 3 moles of Mg can produce 3 moles of hydrogen
Then, 0.00857 moles of Mg must produce 0.00857 moles of H₂
At STP, 1 mol of any gas occupies 22.4L
0.00857 mol . 22.4L / 1mol = 0.190L
The answer is (3), oxidation occurs at the anode and reduction occurs at the cathode. That's because the oxidation reaction can lose electrons and reduction can gain electrons.
Use this formula mass = moles x molar mass.
It should help.
