Question

How many grams of Cl are in 345 g of CaCl2?

Answer 1

Answer on question#38863 - Chemistry - Other

Question

How many grams of Cl are in 345 g of CaCl₂?

Solution:

There are two ways of solution.

1) Molar mass of CaCl₂ equals:

$$M(CaCl_2) = M(Ca) + 2M(Cl) = 40 + 2 \cdot 35.5 = 40 + 71 = 111 \frac{g}{mole}$$

This means that 1 mole of CaCl₂ weighs 111 g. If we have 1 mole of CaCl₂ then 111 g of this compound contains 71 g of Cl. Therefore we can make a proportion:

$$\begin{array}{l} 111 \text{ g of CaCl}_2 - 71 \text{ g of Cl} \\ 345 \text{ g of CaCl}_2 - x \text{ g of Cl} \\ x = \frac{345 \cdot 71}{111} = 220.67 \text{ g} \\ \end{array}$$

**Answer**: $m(Cl) = 220.67 \text{ g}$.

2) 1 mole of CaCl₂ contains 2 moles of Cl (because of formula – there are two Cl-atoms in CaCl₂ molecule). Number of moles of CaCl₂ equals:

$$n = \frac{m}{M}$$

m – Mass of CaCl₂, m = 345 g.

M – Molar mass of CaCl₂, M = 111 g/mole.

$$n = \frac{345}{111} = 3.108 \text{ moles}$$

Therefore we have: $n(Cl) = 2 \cdot 3.108 = 6.216$ moles of CaCl₂. Mass of Cl equals:

$$m(Cl) = n(Cl) \cdot M(Cl) = 6.216 \cdot 35.5 = 220.67 \text{ g}$$

**Answer**: $m(Cl) = 220.67 \text{ g}$.