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>
Sr is the limiting reactant.
Given the reaction equation;
2Sr + O2 (g) → 2SrO
2 moles of Sr reacts with 1 mole of O2
2 moles Sr will react with x mole of O2
x = 2 ×1/2
x = 1 mole of O2
Since we have more O2 than required, it is the reactant in excess, hence Sr is the limiting reactant.
Learn more: brainly.com/question/14225536
Answer:
take the l my gang tfdfhngtyhggggggfggg
The number of C2H5OH in a 3 m solution that contain 4.00kg H2O is calculate as below
M = moles of the solute/Kg of water
that is 3M = moles of solute/ 4 Kg
multiply both side by 4
moles of the solute is therefore = 12 moles
by use of Avogadro law constant
1 mole =6.02 x10^23 molecules
what about 12 moles
=12 moles/1 moles x 6.02 x10^23 = 7.224 x10^24 molecules
Answer:
hope this answer will help you.
