Question

In a reaction container 132 gm of carbon dioxide and 128 gm of calcium oxide are mixed for the formation of CaCO3 gas. What is the limiting reagent and how much CaCO3 is formed in the reaction?

  • CO2 is the limiting reagent and 100 gm of CaCO3 are formed
  • CaO is the limiting reagent and 200 gm of CaCO3 are formed
  • CO2 is the limiting reagent and 200 gm of CaCO3 are formed
  • CaO is the limiting reagent and 150 gm of CaCO3 are formed

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Answer


The correct option is Calcium Oxide (CaO) is the limiting reagent and 200 grams of CaCO$_3$ are formed.

The given reaction is:

$$ \text{CaO} + \text{CO}_2 \rightarrow \text{CaCO}_3 $$

Let's break down the calculations step by step:

  1. Calculating Moles of Carbon Dioxide:

    • Given mass of CO$_2$ = 132 grams.
    • Molar mass of CO$_2$ = 44 g/mol.
    • Number of moles of CO$_2$: $$ \frac{132 , \text{gm}}{44 , \text{g/mol}} = 3 , \text{moles} $$
  2. Calculating Moles of Calcium Oxide:

    • Given mass of CaO = 128 grams.
    • Molar mass of CaO = 64 g/mol.
    • Number of moles of CaO: $$ \frac{128 , \text{gm}}{64 , \text{g/mol}} = 2 , \text{moles} $$
  3. Identifying the Limiting Reagent:

    • According to the reaction stoichiometry, 1 mole of CaO reacts with 1 mole of CO$_2$ to form 1 mole of CaCO$_3$.
    • Since we have 2 moles of CaO and 3 moles of CO$_2$, CaO will run out first. Therefore, CaO is the limiting reagent.
  4. Calculating the Amount of CaCO$_3$ Formed:

    • Since CaO is the limiting reagent, the amount of CaCO$_3$ formed will be determined by the moles of CaO.
    • Moles of CaCO$_3$ formed = moles of CaO initially present = 2 moles.
    • Molar mass of CaCO$_3$ = 100 g/mol.
    • Mass of CaCO$_3$ formed: $$ 2 , \text{moles} \times 100 , \text{g/mol} = 200 , \text{grams} $$

Therefore, the limiting reagent is CaO, and the amount of CaCO$_3$ formed is 200 grams.


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