Calculate the maximum volume in mL of 0.18 M HCl that a tablet containing 340 mg Al(OH)3 and 516 mg Mg(OH)2 would be expected to neutralize. Assume complete neutralization.

171 mL of HCl

Explanation:

The first thing we want to do is consider the reaction between Al(OH)3 and water - as that is the expected reaction that is taking place,

Al(OH)3 + 3HCl → AlCl3 + 3H2O

Knowing this, let's identify the mass of Al(OH)3. Aluminum = 27 g / mol, Oxygen( 3 ) = 16 3 = 48, Hydrogen ( 3 ) = 1 3 = 3 - 27 + 48 + 3 = 78 g / mol. This value is approximated however ( 78 g / mol ), as the molar mass of each substance is rounded as well. Another key thing we need to do here is to convert 340 mg → grams, considering that that unit is a necessity with respect to moles, as you might know - 340 mg = 0.340 g.

Now we can calculate how much moles of HCl will be present in solution, provided we have sufficient information for that,

(0.340 g Al(OH)3) / (78.0036 g / mol Al(OH)3) (3 mol HCl / 1 mol Al(OH)3)

⇒ (.004358773185 g^2 / mol Al(OH)3) (3 HCl / Al(OH)3 )

⇒ .01307632 mol HCl

We can apply this same concept on the reaction of Mg(OH)2 and water, receiving the number of moles of HCl when that takes place. Then we can add the two ( moles of HCl ) and divide by the value " 0.18 mol / L " given to us.

" Mg(OH)2 + 2HCl → MgCl2 + 2H2O "

Molar mass of Mg(OH)2 = 58.3197 g / mol,

516 mg = 0.516 g

(0.516 g Mg(OH)2) / (58.3197 g / mol Mg(OH)2) (2 mol HCl / 1 mol Mg(OH)2)

= .017695564 mol HCL

___________

( .01307632 + .017695564 ) / ( 0.18 M HCl )

= 0.170954911 L

= 171 mL of HCl

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"

periodic trends is an atom that gets larger as the number of electronic shells increase, the radius of atoms increases as you go down a certain group in the periodic table of elements. in general, the size of an atom will decrease as you move from left to right of a certain period.