Answer :
To determine how many grams of [tex]\( HCl \)[/tex] must be used to produce 10.0 grams of chlorine ([tex]\( Cl_2 \)[/tex]), we need to follow a series of steps involving stoichiometry and molar mass calculations. Here is the detailed, step-by-step solution:
1. Write the balanced chemical equation:
[tex]\[ 2 HCl (g) \rightarrow H_2 (g) + Cl_2 (g) \][/tex]
Note that 2 moles of [tex]\( HCl \)[/tex] produce 1 mole of [tex]\( Cl_2 \)[/tex].
2. Find the molar masses:
- Molar mass of [tex]\( HCl \)[/tex]: [tex]\( H \)[/tex] (1.01 g/mol) + [tex]\( Cl \)[/tex] (35.45 g/mol) = 36.46 g/mol
- Molar mass of [tex]\( Cl_2 \)[/tex]: [tex]\( 2 \times Cl \)[/tex] (35.45 g/mol) = 70.90 g/mol
3. Calculate the moles of [tex]\( Cl_2 \)[/tex] produced:
[tex]\[ \text{Moles of } Cl_2 = \frac{\text{mass of } Cl_2}{\text{molar mass of } Cl_2} = \frac{10.0\, \text{g}}{70.90\, \text{g/mol}} \approx 0.141 \text{ mol} \][/tex]
4. Use stoichiometry to find the moles of [tex]\( HCl \)[/tex] required:
According to the balanced equation, the ratio of [tex]\( HCl \)[/tex] to [tex]\( Cl_2 \)[/tex] is 2:1.
[tex]\[ \text{Moles of } HCl = 2 \times \text{moles of } Cl_2 = 2 \times 0.141 = 0.282 \text{ mol} \][/tex]
5. Calculate the mass of [tex]\( HCl \)[/tex] required:
[tex]\[ \text{Mass of } HCl = \text{moles of } HCl \times \text{molar mass of } HCl = 0.282 \times 36.46 \approx 10.285 \text{ g} \][/tex]
Therefore, to produce 10.0 grams of chlorine ([tex]\( Cl_2 \)[/tex]), you must use approximately 10.285 grams of [tex]\( HCl \)[/tex].
1. Write the balanced chemical equation:
[tex]\[ 2 HCl (g) \rightarrow H_2 (g) + Cl_2 (g) \][/tex]
Note that 2 moles of [tex]\( HCl \)[/tex] produce 1 mole of [tex]\( Cl_2 \)[/tex].
2. Find the molar masses:
- Molar mass of [tex]\( HCl \)[/tex]: [tex]\( H \)[/tex] (1.01 g/mol) + [tex]\( Cl \)[/tex] (35.45 g/mol) = 36.46 g/mol
- Molar mass of [tex]\( Cl_2 \)[/tex]: [tex]\( 2 \times Cl \)[/tex] (35.45 g/mol) = 70.90 g/mol
3. Calculate the moles of [tex]\( Cl_2 \)[/tex] produced:
[tex]\[ \text{Moles of } Cl_2 = \frac{\text{mass of } Cl_2}{\text{molar mass of } Cl_2} = \frac{10.0\, \text{g}}{70.90\, \text{g/mol}} \approx 0.141 \text{ mol} \][/tex]
4. Use stoichiometry to find the moles of [tex]\( HCl \)[/tex] required:
According to the balanced equation, the ratio of [tex]\( HCl \)[/tex] to [tex]\( Cl_2 \)[/tex] is 2:1.
[tex]\[ \text{Moles of } HCl = 2 \times \text{moles of } Cl_2 = 2 \times 0.141 = 0.282 \text{ mol} \][/tex]
5. Calculate the mass of [tex]\( HCl \)[/tex] required:
[tex]\[ \text{Mass of } HCl = \text{moles of } HCl \times \text{molar mass of } HCl = 0.282 \times 36.46 \approx 10.285 \text{ g} \][/tex]
Therefore, to produce 10.0 grams of chlorine ([tex]\( Cl_2 \)[/tex]), you must use approximately 10.285 grams of [tex]\( HCl \)[/tex].
