Question

4 moles of $ MgCO{}_3 $ are reacted with $ 6 $ moles of $ HCl $ solution. Find the volume of $ CO{}_2 $ gas produced at STP.
 $ MgCO{}_3 + 2HCl \to MgCl{}_2 + CO{}_2 + H{}_2O $
$ a)67.2lit $
$ b)89.6lit $
$ c)76.3lit $
$ d)22.4lit $

Answer 1

Hint: For determining the volume of carbon dioxide, we should know the given equation is balanced where decomposition reactions take place. We have to find the limiting reagent by stoichiometric coefficient.

Complete Step By Step Answer:
 First let us find the limiting reagent. (A Limiting reagent is used to limit the product in a chemical reaction). We can find the limiting reagent by the following steps;
Balance the given chemical equation.
With the help of molar mass convert everything into moles
Calculate the mole ratio
Compare the calculates and the actual ratio
Use the total amount of the limiting reagent to calculate the amount of product generated.
From the given reaction,
 $ MgCO{}_3 + 2HCl \to MgCl{}_2 + CO{}_2 + H{}_2O $
Given moles is $ 4mole $ and $ 6mole $
Given mole ratio is $ 2:3 $
stoichiometric coefficient is $ 1:2 $
There should be one limiting reagent and to find that we have to divide the given mole by stoichiometric coefficient
 $ MgCO{}_3 $ : $ HCl $ :
 $ \dfrac{4}{1} = 4 $ $ \dfrac{6}{2} = 3 $
Hence $ HCl $ .i.e. hydrochloric acid is the limiting reagent
Moles of HCl divided by $ 2 $ = moles of $ CO{}_2 $ divided by 1
Therefore the total number of produced $ CO{}_2 $ $ = 3 $ moles
Therefore the total volume produced by $ CO{}_2 $ at STP .i.e. Standard temperature and pressure
 $ = 3 \times 22.4 $
 $ = 67.2L $
Hence the total volume produced by $ CO{}_2 $ at STP .i.e. Standard temperature and pressure is $ 67.2L $ .

Note:
$ MgCO{}_3 $ or Magnesium carbonate is a white coloured solid which is an inorganic salt.
Many forms of magnesium exist in the form of minerals. Anhydrous salt also known as magnetite is the most common form of magnesium carbonate.