if 16r of METHANOL are reacted wtih an excess of PBr3, how many grams of CH3Br would be formed if there is a 25% yield?

eqn; CH2OH + PBr3 --> CH3Br + H3PO3

Balance that equation.

Then, how many moles of methanol is 16 grams.

Use the mole ratio to get the moles of CH3Br. Finally, multipy that by .25

To find the number of grams of CH3Br formed, we can follow these steps:

Step 1: Calculate the molar mass of METHANOL (CH2OH).
The molar mass of C = 12.01 g/mol
The molar mass of H = 1.01 g/mol
The molar mass of O = 16.00 g/mol

Total molar mass of CH2OH = (12.01 g/mol x 1) + (1.01 g/mol x 4) + (16.00 g/mol x 1) = 32.05 g/mol

Step 2: Calculate the number of moles of CH2OH.
Given that there are 16r (referred to as "r" to represent relative measure) of CH2OH, we need to convert this into moles.
Using the molar mass of CH2OH:
Number of moles of CH2OH = 16r x (1 mol / 32.05 g/mol)

Step 3: Determine the limiting reactant.
Since we have an excess of PBr3, the limiting reactant is CH2OH.

Step 4: Calculate the number of moles of CH3Br formed.
According to the balanced equation, 1 mol of CH2OH reacts to produce 1 mol of CH3Br.
Therefore, the number of moles of CH3Br formed is equal to the number of moles of CH2OH.

Step 5: Determine the grams of CH3Br formed.
The molar mass of CH3Br is calculated as:
Molar mass of C = 12.01 g/mol
Molar mass of H = 1.01 g/mol
Molar mass of Br = 79.90 g/mol

Total molar mass of CH3Br = (12.01 g/mol x 1) + (1.01 g/mol x 3) + (79.90 g/mol x 1) = 94.94 g/mol

Using the 25% yield, the grams of CH3Br formed is calculated as:
Grams of CH3Br = Number of moles of CH3Br x Molar mass of CH3Br

Now, let's plug in the values and calculate:

Grams of CH2OH = 16r x (1 mol / 32.05 g/mol)
Number of moles of CH3Br = Grams of CH2OH
Grams of CH3Br = (Number of moles of CH3Br) x (Molar mass of CH3Br)

Please provide the value of "r" to proceed with the calculations.

To find the mass of CH3Br formed, we need to follow these steps:

Step 1: Calculate the number of moles of Methanol (CH3OH):
To calculate the number of moles of Methanol, divide the given mass by its molar mass. The molar mass of Methanol (CH3OH) is 32.04 g/mol (12.01 g/mol for carbon + 1.01 g/mol for hydrogen + 16.00 g/mol for oxygen).

Given mass of Methanol = 16g
Molar mass of Methanol = 32.04 g/mol

Number of moles of Methanol = Mass of Methanol / Molar mass of Methanol
= 16g / 32.04 g/mol

Step 2: Determine the limiting reagent using the mole ratio:
The balanced equation shows that the mole ratio between Methanol (CH3OH) and CH3Br is 1:1. So the number of moles of CH3Br formed will be equal to the number of moles of Methanol.

Number of moles of CH3Br formed = Number of moles of Methanol

Step 3: Calculate the mass of CH3Br:
To calculate the mass of CH3Br formed, multiply the number of moles of CH3Br by its molar mass. The molar mass of CH3Br is 94.94 g/mol (12.01 g/mol for carbon + 1.01 g/mol for hydrogen + 79.90 g/mol for bromine).

Molar mass of CH3Br = 94.94 g/mol

Mass of CH3Br formed = Number of moles of CH3Br formed * Molar mass of CH3Br

However, we are given that the yield is 25%. Yield represents the percentage of the theoretical maximum amount of product obtained in a chemical reaction. So, we need to adjust the calculated mass of CH3Br using the given yield.

Adjusted mass of CH3Br formed = Mass of CH3Br formed * Yield

Therefore, we have obtained the answer by the following calculations.