7, 2g of sodium carbonate, Na2Co3.nH2O is strongly heated and a residue of 2.7g are produced..... Calculate the nber of moles of water
2.7/molmassNa2O=2.7/62 = molesNa2O
so you have that number of moles also of Na2CO3.nH20
formula mass Na2CO3.nH20 *moles=mass
or formulamass=7.2/(2.7/62)
=165
so 165=2.7(46+12+48+n18)
or 2.7*18n=165-(2.7*106)
n=2.5
check my math. I was expecting 5.
To calculate the number of moles of water (nH2O) in the given scenario, we need to use the concept of stoichiometry and the molar masses of sodium carbonate (Na2Co3) and water (H2O).
Given:
- Mass of sodium carbonate (Na2Co3) = 2g
- Mass of the residue = 2.7g
First, let's calculate the number of moles of sodium carbonate:
1. Calculate the molar mass of sodium carbonate (Na2Co3):
- Molar mass of Na = 22.99 g/mol
- Molar mass of C = 12.01 g/mol
- Molar mass of O = 16.00 g/mol
- Molar mass of Na2Co3 = (2 * Na) + C + (3 * O) = (2 * 22.99) + 12.01 + (3 * 16.00) = 105.99 g/mol
2. Calculate the number of moles of sodium carbonate:
- Moles of Na2Co3 = Mass / Molar mass = 2g / 105.99 g/mol
Next, let's use the stoichiometry between sodium carbonate (Na2Co3) and water (H2O) to determine the moles of water produced. The balanced equation is:
Na2Co3.nH2O → Na2Co3 + nH2O
Based on the balanced equation, for each mole of sodium carbonate (Na2Co3), an equal number of moles of water (nH2O) are produced.
Therefore, the moles of water (nH2O) produced is equal to the number of moles of sodium carbonate (Na2Co3).
So, the number of moles of water (nH2O) = Moles of Na2Co3
Now, you can substitute the value of the moles of sodium carbonate (Na2Co3) that you calculated in the previous step to find the number of moles of water.