What mass of magnesium oxide results from thermal decomposition of 250*g of magnesium carbonate?

Oct 25, 2016

Approx. $100 \cdot g$ of metal oxide.

Explanation:

With all these sorts of questions, the first thing you need is a stoichiometrically balanced equation:

$M g C {O}_{3} + \Delta \rightarrow M g O + C {O}_{2} \left(g\right) \uparrow$

You simply have to know that under (FIERCE) heating, all metal carbonates decompose to the metal oxide and carbon dioxide as shown. The balanced equation says that each equiv of metal carbonate decomposes to give one equiv magnesium oxide and one equiv of carbon dioxide.

Note that ALL metal carbonates behave this way under heat. If I heated it further, under reducing conditions (i.e. add some silicon for magensium metal), I could probably reduce this oxide to the metal and $S i {O}_{2}$.

So we start with a molar quantity of: $\frac{250 \cdot \cancel{g}}{84.31 \cdot \cancel{g} \cdot m o {l}^{-} 1} = 11.85 \cdot m o l$.

The balanced equation says that I should get $11.85 \cdot m o l$ of $C {O}_{2}$ out if I put $11.85 \cdot m o l$ in. Do you agree? All I am doing is using the balanced equation, which should be the first thing you write.

But if recovery is 85%, then I reasonably get 85%xx11.85*molxx40.30*g*mol^-1=??g

Note that chemists (and you are one!) go to the trouble of writing out the units as an aid in calculation. It is all too easy to mulitply when we should divide, or vice versa, and everybody has done it. In our calculation we had the units: $\cancel{m o l} \times g \cdot \cancel{m o {l}^{-} 1}$, an answer in $\text{grams}$, as we require for a mass. Capisce?

Oct 25, 2016

The balanced equation of thermal decomposition reaction of $M g C {O}_{3}$ is

$\text{ "" "MgCO_3(s)" "stackrel(Delta)->" "MgO(s)" } + C {O}_{2} \left(g\right)$

$\text{Mass"[24+12+3*16]" } \left[24 + 16\right]$

$\text{ "" "[=84]" "" "" "" "" } \left[= 40\right]$

Considring atomic masses as

$M g \to 24 \text{ g/mol }$

$C \to 12 \text{ g/mol }$

$O \to 16 \text{ g/mol}$

As per balanced equation

84g $M g C {O}_{3}$ priduces 40 g MgO

250g $M g C {O}_{3}$ priduces$\frac{40 \cdot 250}{84} g$ MgO

Cinsidering usage of the process as 85% the production of MgO will be
=(40*250)/84gxx85%~~101.2g