5. 2 na + cl₂ → 2 nacl\nwhat volume of chlorine gas, measured at stp, is necessary for the complete reaction…

Explanation:

Step1: Calculate moles of sodium

The molar mass of sodium (Na) is approximately 22.99 g/mol. Using the formula $n=\frac{m}{M}$, where $m$ is mass and $M$ is molar mass. For sodium, $m = 4.81$ g and $M=22.99$ g/mol. So, $n_{Na}=\frac{4.81\ g}{22.99\ g/mol}\approx0.21$ mol.

Step2: Determine mole - ratio from the balanced equation

The balanced chemical equation is $2Na + Cl_2\rightarrow2NaCl$. The mole - ratio of $Na$ to $Cl_2$ is 2:1. So, if $n_{Na} = 0.21$ mol, then $n_{Cl_2}=\frac{0.21\ mol}{2}= 0.105$ mol.

Step3: Calculate volume of chlorine gas at STP

At STP (Standard Temperature and Pressure, $T = 273$ K and $P= 1$ atm), 1 mole of any ideal gas occupies 22.4 L. Using the formula $V=n\times V_m$, where $V_m = 22.4$ L/mol and $n = n_{Cl_2}=0.105$ mol. So, $V_{Cl_2}=0.105\ mol\times22.4\ L/mol = 2.352$ L.

Answer:

2.35 L (rounded to two decimal places)