# How many grams of calcium carbonate are required to prepare for 50.0 grams of calcium oxide ?

## Question:

How many grams of calcium carbonate are required to prepare for {eq}50.0 \ grams {/eq} of calcium oxide ?

## Mole to mole Ratio

The mass of a product or reactant can be determined by multiplying its mole number by its molar mass. This is presented by an equation {eq}mass = (number \ of \ moles) \cdot (Molar\ mass) {/eq}. To take the mole to mole ratio given balanced reaction {eq}aA +bB \rightarrow cC + dD {/eq}, mole ratio is given as {eq}\dfrac{a}{d} {/eq} for a reactant A to product D mole ratio.

Consider the reaction below for solving the mole number of the reactants and products. That is,

{eq}CaCO_3 \to CaO + CO_2 {/eq}

Note that 1 mole of {eq}CaCO_3 {/eq} will produce 1 mole of {eq}CaO {/eq}.

Converting mass to the number of moles of {eq}CaO {/eq} using

{eq}Moles = \dfrac{mass}{Molar\ mass} {/eq}.

By substituting all given values with the molar mass of {eq}CaO {/eq} as {eq}56.08\ \frac{g}{mol} {/eq}. Then,

{eq}Moles = \dfrac{mass}{Molar\ mass}\\ Moles = \dfrac{50.0\ g}{56.08\ \frac{g}{mol}}\\ Moles = 0.892\ moles {/eq}.

Consider finding the mole number of {eq}CaCO_3 {/eq}. That is,

{eq}Moles\ CaCO_3 = (0.892\ moles\ CaO) \cdot \left(\dfrac{1\ mole\ CaCO_3}{1\ mole\ CaO}\right)\\ Moles\ CaCO_3 = 0.892\ mole {/eq}

Now, converting moles to the mass of water with a molar mass of {eq}100.09\ \frac{g}{mol} {/eq}. Therefore,

{eq}mass = (0.892\ moles) \cdot (100.09\ \frac{g}{mol})\\ \boxed{mass = 89.28\ g} {/eq}. 