6 Na + 1 Fe₂O₃ → 3 Na₂O + 6 Fe
<h3>Explanation</h3>
Method One: Refer to electron transfers.
Oxidation states:
- Na: from 0 to +1; loses one electron.
- Fe: from +3 to 0; gains three electrons.
Each mole of Fe₂O₃ contains two Fe atoms and will gain 2 × 3 = 6 electrons during the reaction. It takes 6 moles of Na to supply all those electrons.
6 Na + 1 Fe₂O₃ → ? Na₂O + ? Fe
- There are two moles of Na atoms in each mole of Na₂O. 6 moles of Na will make 3 moles of Na₂O.
- There are two moles of Fe atoms in each mole of Fe₂O₃. 1 mole of Fe₂O₃ will make 2 moles of Fe.
6 Na + 1 Fe₂O₃ → 3 Na₂O + 2 Fe
Method Two: Atoms conserve.
Fe₂O₃ has the largest number of atoms among one mole of all four species in this reaction. Assume <em>one</em> as its coefficient.
? Na + <em>1</em> Fe₂O₃ → ? Na₂O + ? Fe
There are two moles of Fe atoms and three moles of O atoms in each mol of Fe₂O₃. One mole of Fe₂O₃ contains two moles of Fe and three moles of O. There are one mole of O atom in every mole of Na₂O. Three moles of O will go to three moles of Na₂O.
? Na + <em>1</em> Fe₂O₃ → <em>3</em> Na₂O + <em>2</em> Fe
Each mole of Na₂O contains two moles of Na. Three moles of Na₂O will contain six moles of Na.
<em>6</em> Na + <em>1</em> Fe₂O₃ → <em>3</em> Na₂O + <em>2</em> Fe
Simplify the coefficients. All coefficients in this equation are now full number and relatively prime. Hence the equation is balanced.
6 Na + 1 Fe₂O₃ → 3 Na₂O + 2 Fe
I believe the awnser is a. universal solvent.
To fully understand the problem, we use the ICE table to identify the concentration of the species. We calculate as follows:
Ka = 2.0 x 10^-9 = [H+][OBr-] / [HOBr]
HOBr = 0.50 M
KOBr = 0.30 M = OBr-
<span> HOBr + H2O <-> H+ + OBr- </span>
<span>I 0.50 - 0 0.30 </span>
<span>C -x x x
</span>---------------------------------------------
<span>E(0.50-x) x (0.30+x) </span>
<span>Assuming that the value of x is small as compared to 0.30 and 0.50 </span>
<span>Ka = 2.0 x 10^-9 = x (0.30) / 0.50) </span>
<span>x = 3.33 x 10^-9 = H+</span>
pH = 8.48
