1) How many grams of CH4 (MM = 16.043) are needed to react with 50.0g of CL2 (MM=70.91)?
2) What is the molarity of a solution prepared by dissolving 20.0 g of methanol, CH3OH, oara sufficient water to prepare 500 mL of solution? Show calculations
1. You need an equation for this since CH4 and Cl2 react to give several products.
2. mols CH3OH = grams/molar mass = ?
Then M = mols CH3OH/L solution
1) To determine the grams of CH4 needed to react with 50.0g of Cl2, we need to first balance the chemical equation:
CH4 + Cl2 -> CH3Cl + HCl
From the balanced equation, we can see that 1 mole of CH4 reacts with 1 mole of Cl2. We need to find the moles of Cl2 present:
Moles of Cl2 = Mass (g) / Molar Mass (g/mol)
Moles of Cl2 = 50.0g / 70.91 g/mol
Now, since the ratio of moles of CH4 to Cl2 is 1:1, the moles of CH4 required is also equal to Moles of Cl2. Therefore, the grams of CH4 needed can be calculated as follows:
Grams of CH4 = Moles of CH4 x Molar Mass of CH4
Grams of CH4 = (50.0g / 70.91 g/mol) x 16.043 g/mol
2) To find the molarity of a solution prepared by dissolving 20.0g of methanol, CH3OH, in sufficient water to prepare 500 mL (0.500 L) of solution, we need to first calculate the moles of methanol:
Moles of CH3OH = Mass (g) / Molar Mass (g/mol)
Moles of CH3OH = 20.0g / 32.04 g/mol
Next, we can calculate the molarity using the formula:
Molarity (M) = Moles of solute / Volume of solution (L)
Molarity (M) = (Moles of CH3OH / 0.500 L)
Please note that water does not affect the molarity calculation since it does not dissociate in solution. Hence, the volume of water does not need to be considered in this case.