using this equation: CH4+2Cl2-> CH2Cl2+2HCl. If 6.00g of Ch4 are mixed with 6.00g of Cl2, how much HCl can be made?

See

http://www.jiskha.com/display.cgi?id=1269788658

To determine how much HCl can be made, we need to find the limiting reagent and then calculate the amount of HCl that can be produced based on the amount of the limiting reagent.

1. Start by calculating the number of moles for each reactant:
- Moles of CH4 = mass / molar mass = 6.00g / 16.04 g/mol (molar mass of CH4)
- Moles of Cl2 = mass / molar mass = 6.00g / 70.90 g/mol (molar mass of Cl2)

2. Use the stoichiometric coefficients from the balanced equation to determine the mole ratio between CH4 and HCl.
From the balanced equation: CH4 + 2Cl2 -> CH2Cl2 + 2HCl
The mole ratio between CH4 and HCl is 1:2.

3. Next, determine the limiting reagent.
- Calculate the number of moles of HCl that can be produced from CH4 (theoretically):
Moles of HCl = Moles of CH4 x (2 moles of HCl / 1 mole of CH4)
- Calculate the number of moles of HCl that can be produced from Cl2 (theoretically):
Moles of HCl = Moles of Cl2 x (2 moles of HCl / 2 moles of Cl2)

The limiting reagent is the reactant that produces the least amount of HCl. So, the limiting reagent is Cl2 in this case, as it produces fewer moles of HCl compared to CH4.

4. Now, calculate the actual amount of HCl that can be produced based on the limiting reagent.
- Moles of Cl2 used = moles of Cl2 (calculated in step 1)
- Moles of HCl produced = moles of Cl2 used x (2 moles of HCl / 2 moles of Cl2)
- Mass of HCl produced = Moles of HCl produced x molar mass of HCl

Finally, you can calculate the amount of HCl produced in grams by multiplying the number of moles of HCl by its molar mass.

Note: Make sure to round the final answer to an appropriate number of significant figures.