How many grams of water would it take to quench 1 g of phenylmagnesium bromide?
Write the equation and balance it.
phMgBr + HOH ==>
Convert 1 g phMgBr to moles. moles = grams/ molar mass.
Using the coefficients in the balanced equation, convert moles phMgBr to moles water.
Now convert moles water to grams. grams = mols water x molar mass water.
.099 grams
To determine the amount of water needed to quench 1 gram of phenylmagnesium bromide, we need to consider the stoichiometry of the reaction between phenylmagnesium bromide and water.
The balanced chemical equation for the reaction between phenylmagnesium bromide and water is as follows:
C6H5MgBr + H2O → C6H6 + MgBr(OH)
From the equation, we can see that 1 mole of phenylmagnesium bromide (C6H5MgBr) reacts with 1 mole of water (H2O) to produce 1 mole of benzene (C6H6) and 1 mole of magnesium bromide hydroxide (MgBr(OH)).
First, let's calculate the molar mass of phenylmagnesium bromide (C6H5MgBr):
- Carbon (C): 12.01 g/mol
- Hydrogen (H): 1.01 g/mol
- Magnesium (Mg): 24.31 g/mol
- Bromine (Br): 79.90 g/mol
Molar mass of C6H5MgBr:
(6 * 12.01 g/mol) + (5 * 1.01 g/mol) + 24.31 g/mol + 79.90 g/mol = 216.28 g/mol
Now, we can calculate the amount of water needed using the molar ratio between phenylmagnesium bromide and water.
Molar ratio between C6H5MgBr and H2O:
1 mole C6H5MgBr : 1 mole H2O
Since the molar mass of phenylmagnesium bromide is 216.28 g/mol, we can calculate the amount of water needed in grams:
(1 g C6H5MgBr) x (1 mol H2O / 216.28 g C6H5MgBr) = 0.00462 g H2O
Therefore, approximately 0.00462 grams of water would be needed to quench 1 gram of phenylmagnesium bromide.