In a chemical reaction, exactly 2 mol of substance A react to produce exactly 3 mol of substance B.
How many molecules of substance B are produced when 26.5 g of substance A reacts? The molar mass of substance A is 21.9 g/mol.
Step 1: Convert the mass of A to moles.
Step 2: Convert the number of moles of A to the number of moles of B.
Step 3: Convert the number of moles of B to molecules of B.
So what is your problem? The directins are there. The equation is
2A ==> 3B
To convert grams to mols remember mols = grams/molar mass.
To convert mols to molecules remember 1 mol of anything contains 6.02E23 of those "anythings".
Step 1: Convert the mass of A to moles.
To convert the mass of substance A to moles, we use the formula:
moles = mass / molar mass
In this case, the mass of substance A is given as 26.5 g, and the molar mass of A is given as 21.9 g/mol. Plugging these values into the formula, we get:
moles of A = 26.5 g / 21.9 g/mol
Calculating this, we find that there are approximately 1.21 moles of substance A.
Step 2: Convert the number of moles of A to the number of moles of B.
From the given information, we know that 2 moles of A react to produce 3 moles of B. So, we can set up a proportion using the ratio of moles of A to moles of B:
2 moles A / 3 moles B = 1.21 moles A / x moles B
Solving for x, we can cross-multiply and divide to get:
x moles B = (3 moles B * 1.21 moles A) / 2 moles A
Calculating this, we find that there are approximately 1.81 moles of substance B.
Step 3: Convert the number of moles of B to molecules of B.
To convert moles of substance B to molecules, we use Avogadro's number, which states that 1 mole of any substance contains 6.022 x 10^23 molecules.
Using this, we can calculate the number of molecules of substance B:
molecules B = moles B * Avogadro's number
Plugging in the value of moles B as 1.81, we get:
molecules B = 1.81 moles B * 6.022 x 10^23 molecules/mol
Calculating this, we find that there are approximately 1.09 x 10^24 molecules of substance B when 26.5 g of substance A reacts.