Answer:
The empirical formula for the compound is Na2O
Explanation:
Data obtained from the question include:
Sodium (Na) = 74.2g
Oxygen (O) = 25.8g
We can obtain the empirical formula for the compound as follow:
First, divide the above by their individual molar mass as shown below:
Na = 74.2/23 = 3.226
O = 25.8/16 = 1.613
Next, divide the above by the smallest number
Na = 3.226/1.613 = 2
O = 1.613/1.613 = 1
Therefore, the empirical formula is:
Na2O
Answer:
27 g
Explanation:
M(C6H12O6) = 6*12 + 12*1 + 6*16 = 180 g/mol
100 mL = 0.1 L solution
1.5 M = 1.5 mol/L
1.5 mol/L * 0.1 L = 0.15 mol C6H12O6
0.15 mol * 180 g/1 mol = 27 g C6H12O6
The term formula units means molecules.
Then, what you are looking for is the mass in 4.59*10^24 molecules.
The procedure involves to convert the 4.59 * 10^24 molecules into moles and use the molar mass of the sodium chloride.
1) Number of moles = 4.59 * 10^24 molecules / (6.02 * 10^23 molecules/mol) = 7.62 mol
2) Molar mass of NaCl = 22.99 g/mol + 35.45 g/mol = 58.44 g/mol
3) mass of NaCl = molar mass * number of moles = 58.44 g/mol * 7.62 mol = 445.31 g of NaCl
Answer: 445.31 g of NaCl.
The balanced chemical reaction is written as :
Na2CO3<span> + 2HCl === 2NaCl + H2O + CO2
</span>
We are given the amount of NaCl to be produced from the reaction. This will be the starting point for the calculations. We do as follows:
120 g NaCl ( 1 mol / 58.44 g) ( 1 mol Na2CO3 / 2 mol NaCl)( 105.99 g / 1 mol ) = 1108.82 g Na2CO3 needed
