Respuesta :
First, we have to calculate the molecular weights of each molecule:
[tex]\begin{gathered} Na_2CO_3\text{ : 23*2+12+16*3= 106 g/mol} \\ H_2O\text{ : 1*2+16= 18 g/mol} \end{gathered}[/tex]Then, we have to calculate the number of grams of water. We can calculate them because the process of evaporation lets us know the water amount that was retired:
[tex]g\text{ H}_2O\text{ = 35.40 g - 30.2 g=5.2 g H}_2O[/tex]Then, we're gonna convert the grams of sodium carbonate alone (30.2 g) and the grams of water to moles:
[tex]\begin{gathered} 30.2\text{ g Na}_2CO_3\text{ * }\frac{1\text{ mol}}{106\text{ g}}=\text{ 0.2849 mol Na}_2CO_3\text{ }\approx0.3\text{ mol Na}_2CO_3 \\ \\ 5.2\text{ g H}_2O\text{ * }\frac{1\text{ mol}}{18\text{ g}}=\text{ 0.288 mol H}_2O\text{ }\approx\text{ 0.3 mol H}_2O \end{gathered}[/tex]It means that the mole relation is 1:1 approx, as it is the same amount for both. Then, the formula is going to be:
[tex]Na_2CO_3\text{ . 1H}_2O[/tex]It means that the answer is A.
