Question

How many liters of chlorine gas can react with 56.0 grams of calcium metal at standard temperature and pressure? Show all of the work used to find your answer.
Ca + Cl2 CaCl2

Answer 1

Answer:

31.36 L of chlorine gas will react with 56 g of calcium.

Explanation:

$Ca + Cl_2\rightarrow CaCl_2$

Mass of calcium = 56 g

Moles of calcium gas = $\frac{56 g}{40 g/mol}=1.4 moles$

According to reaction, 1 mole of calcium react with 1 mole of chlorine gas.

Then 1.4 mol of calcium wiull react with :

$\frac{1}{1}\times 1.4 = 1.4 mol$ chlorine gas

At standard temperature and pressure. 1 mol of gas occupies 22.4 L of volume.

Then 4.1 moles of chlorine gas will occupy :

$22.4 L\tmes 4.1 =31.36 L$

31.36 L of chlorine gas will react with 56 g of calcium.

Answer 2

The balanced chemical reaction is:

Ca + Cl2 =  CaCl2

We are given the amount of calcium metal to be used for this reaction. This will be the starting point for the calculations.

56 g Ca ( 1 mol Ca / 40.08 g Ca) (1 mol Cl2 / 1 mol Ca) ( 22.414 L Cl2 / 1 mol Cl2 ) = 31.32 L Cl2 gas produced from the reaction