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?

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

4.68 L

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

Trees / Forests / Rain forests

Explanation:

Trees remove CO₂ from air because they need it for their cellular respiration.