Question

What volume of chlorine gas at 45.3 c is needed to react with 14.2g of sodium to form nacl at 1.72 atm?

Answer 1

4.68 L

Step-by-step explanation:

Fist, we need to know the balanced reaction:

Cl2 + 2 Na -> 2NaCl

To know the volume of Cl2, we should use the ideal gas equation PV = nRT. Where P= 1.72 atm, R = 0.082 atm.L/(mol.K) which is the constant of the gases, and T = 45.3 + 273 = 318.3 K.

So, we need the number of moles n. By the stoichiometry of the reaction, for 1 mol of Cl2, 2 moles of Na are needed. 1 mol of Cl2 has 71 g ( MM Cl = 35.5 g/mol), and 2 moles of Na have 46 g ( MM of Na = 23 g/mol). To calculate the mass of chlorine gas, we use a simple three rule:

71 g of Cl2 --------------------- 46 of Na

X --------------------- 14.2 g

46X = 1,008.2

X = 21.92 g

As the molar mass of Cl2 is 71 g/mol, the number of moles are:

n = m/MM

n = 21.92/71

n = 0.3087 moles

So, the volume would be:

PV = nRT

1.72V = 0.3087x0.082x318.3

1.72V = 8.0572

V = 4.68 L

Answer 2

To determine the volume of chlorine gas that is needed, we need to first know the reaction. It is expressed as follows:

2Na + Cl2 = 2NaCl

Then, from the amount of NaCl to be produced, we calculate the moles of Cl2 needed.

14.2 g NaCl ( 1 mol / 58.44 g ) ( 1 mol Cl2 / 2 NaCl ) = 0.1215 mol Cl2

To determine the volume of the gas, we need an equation that would relate the number of moles to volume. There are a number of equation available but some are complex equations so we assume that it is an ideal gas. We use the equation:

PV = nRT
V = nRT / P
V = 0.1215 mol ( 0.08205 L-atm / mol-K ) (45.3 +273.15 K) / 1.72 atm
V = 1.85 L Cl2 gas