In the chlor-alkali process, the mass of NaCl(s) required to produce 0.793 kg of chlorine gas is _____ kg.
The process reaction is:
2 NaCl + 2 H2O -> 2NaOH + H2 + Cl2
Convert 793 g of Cl2 to moles. It is about 11.17 moles.
You will need twice as many moles of NaCl. Convert that number to mass of NaCl, in the usual manner.
http://en.wikipedia.org/wiki/Chloralkali_process
To find the mass of NaCl(s) required to produce chlorine gas in the chlor-alkali process, we need to determine the stoichiometry of the reaction between NaCl and chlorine gas. Let's write the balanced chemical equation for this process:
2 NaCl(s) + 2 H2O(l) → Cl2(g) + H2(g) + 2 NaOH(aq)
From the balanced equation, we can see that 2 moles of NaCl react to produce 1 mole of Cl2 gas.
Now, we need to convert the given mass of chlorine gas (0.793 kg) to moles using its molar mass. The molar mass of chlorine gas (Cl2) is approximately 70.906 g/mol.
0.793 kg Cl2 × (1000 g / 1 kg) × (1 mol Cl2 / 70.906 g) = 11.18 mol Cl2
Since the stoichiometry of the reaction is 2 moles of NaCl to 1 mole of Cl2 gas, we can conclude that:
11.18 mol Cl2 × (2 mol NaCl / 1 mol Cl2) = 22.36 mol NaCl
Finally, we convert the moles of NaCl to grams by multiplying by its molar mass. The molar mass of NaCl is approximately 58.44 g/mol.
22.36 mol NaCl × 58.44 g/mol = 1306 g
Therefore, the mass of NaCl(s) required to produce 0.793 kg of chlorine gas is approximately 1306 grams.