Question

question 3 (2 points)

use the molar mass of al₂(so₄)₃ calculated in the previous problem to calculate how many moles of aluminum sulfate are in 10.0 g.
the previous problem gave 342.15 g/mol for aluminum sulfate, and this is a direct proportion and a conversion factor. use units to set up the problem.

(10.0 g as)(\frac{\\_\\_1\\_\\_mol as}{\\_\\_3\\_\\_g as}) = \\_\\_2\\_\\_ mol as

a. 0 b. 1 c. 2 d. 3 e. 4 f. 342.15
g. 10.0 h. 56.1 i. 0.0292 j. 3420

Explanation:

Step1: Determine the conversion factor

The conversion factor is \(\frac{1\space mol\space AS}{342.15\space g\space AS}\) (since molar mass is \(342.15\space g/mol\)).

Step2: Calculate the number of moles

$$n=(10.0\space g\space AS)\times\frac{1\space mol\space AS}{342.15\space g\space AS}$$

$$n=\frac{10.0}{342.15}\space mol\space AS\approx 0.0292\space mol\space AS$$

Answer:

1. B. 1
2. I. 0.0292
3. F. 342.15