Answer:
6Br⁻ + XeO₃ + 6H⁺ → 3Br₂ + Xe + 3H₂O
Explanation:
First, we need to write the half-reactions:
2Br⁻ → Br₂ + 2e⁻ Oxidation -Balanced yet-
XeO₃ → Xe Reduction
To balance the reduction in acidic aqueous solution we need to add waters in the other side of the reaction as oxygens are present:
XeO₃ → Xe + 3H₂O
And H⁺ as hydrogens from water we have:
XeO₃ + 6H⁺ → Xe + 3H₂O
To balance the charge:
<h3>XeO₃ + 6H⁺ + 6e⁻ → Xe + 3H₂O Reduction -Balanced-</h3><h3 />
To cancel out the electrons of both half-reaction we need to multiply oxidation 3 times:
6Br⁻ → 3Br₂ + 6e⁻
XeO₃ + 6H⁺ + 6e⁻ → Xe + 3H₂O
And the balanced reaction in acidic aqueous solution is the sum of both half-reactions:
<h3>6Br⁻ + XeO₃ + 6H⁺ → 3Br₂ + Xe + 3H₂O </h3>
<span>Answer: 56.6 moles
Explanation:
28.3 moles of Pb would produce twice as much moles as Ag.
28.3 X (2moles Ag/ 1 mol Pb) = 56.6 moles of Ag.</span>
1 kpa = 0.0098692327 atm so just multiply that by 45.6
2(NH4)3PO4 (aq) + 3Ni(NO3)2(aq) ------> Ni3(PO4)2(s) + 6NH4NO3 (aq)
Ni3(PO4)2 is a precipitate.
