Chemistry Help (Molarity)?

Posted on 2016-05-11 18:02:00 in Chemistry by ESTELLA

A 10.0 g piece of pure aluminum is placed in 75.0 mL of 0.54 M hydrochloric acid at STP condition. They react as follows: 

2Al 6HCl → 3H2(g) 2AlCl3 

Calculate the following: 

a. Volume, in liters, of hydrogen gas. 

b. Molarity of Al 3. (Assume 75.0 mL solution.) 

c. Molarity of Cl–. (Assume 75.0 mL solution.) 

I think I figured out part A.... 
10.0 g Al x 1 mol Al/ 26.98 g Al = 0,371 mol AL 

0.371 mol Al x 3 mol H2/ 2 mol Al = 0.557 mol H2 

0.557 mol H2 x 22.4 L / 1 mol H2= 12.5 L H2 

Is this right? If not can someone please explain? Part B and C I have no clue, help please.

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