a. How many moles of Na2Co3 were present in the 15mL sample?
b. How many grams of Na2CO3 were present in the 15mL sample?
c.How many grams of Na2CO3 are present in the 575L of the Na2CO3 solution?
16 years ago
16 years ago
DrBoB222 you responds in unknown
16 years ago
Please review what you have posted. You have no data upon which to base an answer. Questions but no data.
16 years ago
This is a Lab Stoichiometry and Gravimetric Analaysis
Balance equation:
Na2CO3 + CaCl2 ---> CaCO3 + 2NaCl
Precipitate(?): Not Sure
Na2CO3 + CaCl2
Calculate the mass of the dry precipitate and the number of moles of precipitate produced ine the reactation?
Empirical Formula(?):
?
16 years ago
answered above.
11 months ago
To find the answers to these questions, we need some information:
1. The molar mass of Na2CO3, which is 105.99 g/mol.
2. The concentration of Na2CO3 in the solution, which is not provided. We'll assume it is given in moles per liter (mol/L).
a. To find the number of moles of Na2CO3 present in the 15 mL sample, we need to know the concentration of Na2CO3 in moles per liter. Let's assume it is C mol/L.
Number of moles of Na2CO3 = Concentration x Volume
Number of moles of Na2CO3 = C mol/L x 15 mL x (1 L / 1000 mL)
b. To find the number of grams of Na2CO3 present in the 15 mL sample, we can use the molar mass of Na2CO3.
Number of grams of Na2CO3 = Number of moles x Molar mass
Number of grams of Na2CO3 = C mol/L x 15 mL x (1 L / 1000 mL) x 105.99 g/mol
c. To find the number of grams of Na2CO3 present in the 575 L solution, we need to know the concentration of Na2CO3 in moles per liter. Let's assume it is C' mol/L.
Number of grams of Na2CO3 = Concentration x Volume x Molar mass
Number of grams of Na2CO3 = C' mol/L x 575 L x 105.99 g/mol
Please provide the concentration of Na2CO3 in mol/L for a more accurate calculation.