# How many milliliters of 0.306 M HCl are needed to react with 51.0 g of CaCO3?

Jul 1, 2017

3330 mL

#### Explanation:

From the chemical equation, we can see that for every mole of $C a C {O}_{3}$, 2 moles of HCl react. This means that if you want them to react exactly in proportion, the number of moles of HCl needed is double that of $C a C {O}_{3}$.

$n \left(H C l\right) = 2 \cdot n \left(C a C {O}_{3}\right)$

The molar mass (MM) of calcium carbonate is:

$M M \left(C a C {O}_{3}\right) = 40.1 + 12.0 + 3 \cdot 16.0 = 100.1 \text{ } \frac{g}{m o l}$

The number of moles of calcium carbonate can now be calculated:

$n \left(C a C {O}_{3}\right) = \frac{m a s s}{M M} = \frac{51.0}{100.1} = 0.509 \text{ } m o l$

So, now we double the answer to get the amount of HCl needed:

$n \left(H C l\right) = 2 \cdot n \left(C a C {O}_{3}\right) = 2 \cdot 0.509 = 1.02 \text{ } m o l$

Rearranging the concentration formula we can get the volume needed in litres:

$C = \frac{n}{V} \Rightarrow V = \frac{n}{C}$

$V = \frac{1.02}{0.306} = 3.33 \text{ } L$

Convert to mL:

$V = 3.33 \cdot {10}^{3} \text{ } m L$