Quote: Originally posted by Antiswat ![]() |
Quote: Originally posted by Antiswat ![]() |
<sub>2</sub> is 13.181% Mg by mass.
<a href="https://en.wikipedia.org/wiki/Dolostone" target="_blanK">Dolostone</a> <img src="../scipics/_wiki.png" />, a common rock
composed of primarily dolomite, would be slightly less than 13.181 % Mg by mass. Assuming that the contents are powdered dolostone with a purity of
91% dolomite<sup>1</sup>, the product would contain 12 % Mg by mass. <em>Your <a
href="https://en.wikipedia.org/wiki/Agricultural_lime" target="_blank">agricultural lime</a> <img src="../scipics/_wiki.png" />
('dolomit kalk') is</em>—<a href="https://en.wikipedia.org/wiki/Occam's_razor" target="_blank">beyond a reasonable
doubt</a> <img src="../scipics/_wiki.png" />—<em>powdered dolostone in a bag.</em> The remaining 9 % is likely
quartz grains and other minerals commonly present in sedimentary rock, and possibly a little water of hydration.
<sub>2</sub> × 100 % ≈ 13.181 % Mg by mass
<sub>2</sub>
<sub>2</sub> × 13.181 %
Mg = 11.99471 % Mg –(2 s.f.)→ 12 % Mg by mass Quote: Originally posted by Antiswat ![]() |
<sub>2</sub>] during the crude analysis. The actual minerals that are attributed to the percent dolomite would include dolomite,
calcite (CaCO<sub>3</sub>
, magnesite (MgCO<sub>3</sub>
, ankerite (Ca(Fe,Mg,Mn)(CO<sub>3</sub>
<sub>2</sub>
, etc.
is
<em>Kreide</em> in German (<a href="http://books.google.com/books?id=0WRIAAAAMAAJ" target="_blank">ref.</a> <img
src="../scipics/_ext.png" />
and <em>Kridt</em> in Danish.
, and ca. 13 kg of material that acts like calcium carbonate during the crude chemical analysis and use as a
fertilizer. I can't think of a clearer way of explaining this; so if you're still having difficulty, perhaps someone else would be kind enough to
provide a simpler perspective.
<sub>2</sub> + 2
H<sub>2</sub>SO<sub>4</sub>(aq) → CaSO<sub>4</sub>(aq) + MgSO<sub>4</sub>(aq) + 2
H<sub>2</sub>O + 2 CO<sub>2</sub>(g)
<sub>2</sub> ÷ 184.401 g/mol
CaMg(CO<sub>3</sub>
<sub>2</sub> × 100 % =
CaMg(CO<sub>3</sub>
<sub>2</sub> is 65.09 %
CO<sub>3</sub><sup>2-</sup>, by mass
<sub>2</sub> × 65.09 % CO<sub>3</sub><sup>2-</sup> × 73.5 %
CO<sub>2</sub>(g) ≈ 11 kg CO<sub>2</sub>(g) / bag (<em>assuming no other carbonates present!</em>