Question

A student used the procedure in this experiment to determine the empirical formula of aluminum chloride. In the student's experiment, a sample of aluminum (Al), weighing 1.500 g, reacted completely with an excess of aqueous HCl to produce 7.431 g of aluminum chloride. (1) Determine the percent Al in the aluminum chloride formed during the reaction. (2) Determine the percent Cl in the aluminum chloride formed during the reaction.

Answer 1

The percent Al in the aluminum chloride formed during the reaction is 20.21%, and the percent Cl is 79.96%.

Step-by-step explanation:

To determine the percent Al in the aluminum chloride formed during the reaction, we can use the formula:

Percent Al = (mass of Al / mass of AlCl3 formed) * 100%

In this case, the mass of Al is 1.500 g and the mass of AlCl3 formed is 7.431 g. Plugging in these values into the formula, we get:

Percent Al = (1.500 g / 7.431 g) * 100% = 20.21%

To determine the percent Cl in the aluminum chloride formed during the reaction, we can use the formula:

Percent Cl = (mass of Cl / mass of AlCl3 formed) * 100%

In this case, the mass of Cl can be found by subtracting the mass of Al from the mass of AlCl3 formed. The mass of Cl is therefore 7.431 g - 1.500 g = 5.931 g. Plugging in these values into the formula, we get:

Percent Cl = (5.931 g / 7.431 g) * 100% = 79.96%