The decomposition of a certain mass of $CaCO_3$ gave $11.2\, dm^3$ of $CO_2$ gas at $STP$. The mass of $KOH$ required to completely neutralise the gas is
- 56 g
- 28 g
- 42 g
- 20 g
The Correct Option is B
Solution and Explanation
$\underset{56\,g}{KOH}+\underset{44\,g}{CO_2}\to KHCO_3$
KOH required for complete neutralisation requires of $22 \,g$ of $CO _{2}$ is
$=\frac{56}{44} \times 22=28 \,g$
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Concepts Used:
Stoichiometry
Stoichiometry is founded on the law of conservation of mass where the total mass of the reactants equals the total mass of the products, leading to the insight that the relations among quantities of reactants and products typically form a ratio of positive integers. This means that if the amounts of the separate reactants are known, then the amount of the product can be calculated. Conversely, if one reactant has a known quantity and the quantity of the products can be empirically determined, then the amount of the other reactants can also be calculated.
Stoichiometry helps us determine how much substance is needed or is present. Things that can be measured are;
- Reactants and Products mass
- Molecular weight
- Chemical equations
- Formulas
Stoichiometric Coefficient
The Stoichiometric coefficient of any given component is the number of molecules and/or formula units that participate in the reaction as written.
Mole Ratios
The mass of one mole of a substance in grams is called molar mass. The molar mass of one mole of a substance is numerically equal to the atomic/molecular formula mass.