<span>The region(s) of the periodic table which are
made up of elements that can adopt both positive and negative oxidation numbers
are the “non-metal” region. As we can see on the periodic table, the elements situated
at the right side of the table have two oxidation states, one positive and the
other a negative. </span>
<span><span>m1</span>Δ<span>T1</span>+<span>m2</span>Δ<span>T2</span>=0</span>
<span><span>m1</span><span>(<span>Tf</span>l–l<span>T<span>∘1</span></span>)</span>+<span>m2</span><span>(<span>Tf</span>l–l<span>T<span>∘2</span></span>)</span>=0</span>
<span>50.0g×<span>(<span>Tf</span>l–l25.0 °C)</span>+23.0g×<span>(<span>Tf</span>l–l57.0 °C)</span>=0</span>
<span>50.0<span>Tf</span>−1250 °C+23.0<span>Tf</span> – 1311 °C=0</span>
<span>73.0<span>Tf</span>=2561 °C</span>
<span><span>Tf</span>=<span>2561 °C73.0</span>=<span>35.1 °C</span></span>
The chemical reaction would be as follows:
<span>2Na + S → Na2S
We are given the amount of the reactants to be used in the reaction. We use these to calculate the amount of product. We do as follows:
45.3 g Na ( 1 mol / 22.99 g ) = 1.97 mol Na
105 g S ( 1 mol / 32.06 g ) = 3.28 mol S
The limiting reactant would be Na. We calculate as follows:
1.97 mol Na ( 1 mol Na2S / 2 mol Na ) (78.04 g / mol ) = 76.87 g Na2S produced</span>
