contestada

A 1.2048 gram impure sample of Na2CO3 is dissolved and allow to react with a solution of CaCl2. The resulting CaCO3 after precipitation, filtration and drying was found to weigh 1.0262g. Assuming that the impurities do not contribute to the weight of the precipitate, calculate the percentage purity of Na2CO3.

Respuesta :

pstnonsonjoku

From the calculation, the percentage purity of the material is 90.5%.

What is a chemical reaction?

A chemical reaction is said to occur when there is a combination between two or more substances.

In this case, the reaction is; Na2CO3 + CaCl2 ----> 2NaCl + CaCO3.

Number of moles of CaCO3  = 1.0262g/100 g/mol = 0.0103 moles

Since the reaction is 1:1, 0.0103 moles of Na2CO3  reacted.

Mass of Na2CO3  = 0.0103 moles  * 106 g/mol = 1.09 g

Percent by mass of Na2CO3  = 1.09 g/1.2048 g * 100/1 = 90.5%

Learn more about percentage purity:https://brainly.com/question/17836636

#SPJ1

Otras preguntas