A chemist requires 0.780 mol Na2CO3 for a reaction. How many grams does this correspond to?
just get the molar mass and use multiplication by one, meaning it would be mol/1 x grams(from molar mass)/moles(1)of the compound.
I think it is easier this way.
grams = mols x molar mass.
To convert from moles to grams, we need to use the molar mass of Na2CO3. The molar mass of Na2CO3 can be calculated by summing the atomic masses of its elements.
The atomic masses of the elements are:
Na: 22.99 g/mol
C: 12.01 g/mol
O: 16.00 g/mol
The molar mass of Na2CO3 is:
(2 * Na) + C + (3 * O) = (2 * 22.99 g/mol) + 12.01 g/mol + (3 * 16.00 g/mol) = 105.99 g/mol
Now, we can use this molar mass to calculate the grams of Na2CO3 required for the reaction.
0.780 mol Na2CO3 * 105.99 g/mol = 82.6152 g
Therefore, 0.780 mol of Na2CO3 corresponds to 82.6152 grams.
To determine the number of grams of Na2CO3 needed for the reaction, you can use the molar mass and the given number of moles.
The molar mass of Na2CO3 can be calculated by finding the sum of the atomic masses of its constituent elements from the periodic table. Na has an atomic mass of 22.99 g/mol, C has an atomic mass of 12.01 g/mol, and O has an atomic mass of 16.00 g/mol.
Molar mass of Na2CO3 = (2 * Na) + C + (3 * O)
= (2 * 22.99 g/mol) + 12.01 g/mol + (3 * 16.00 g/mol)
= 46.00 g/mol + 12.01 g/mol + 48.00 g/mol
= 106.01 g/mol
Now, we can calculate the number of grams using the given number of moles:
Number of moles = 0.780 mol
Number of grams = Number of moles * Molar mass
= 0.780 mol * 106.01 g/mol
= 81.780 g
Therefore, 0.780 mol of Na2CO3 corresponds to 81.780 grams.