For the reaction HNO3 + Mg(OH)2→ Mg(NO3)2 + H2O, how many grams of magnesium nitrate are produced from 8.00 mol of nitric acid, HNO3? Balance the equation

tha mass of magnesium nitrate is 592g

Explanation:

from a balanced chemical equation

2HNO3 + Mg(OH)2→ Mg(NO3)2 + 2H2O

2 mol of nitric acid is equivalent to 1 mol of magnesium nitrate. then 8 mol of nitric acid will be equivalent to 4 mol of magnesium nitrate.

2HNO3 + Mg(OH)2 --> 2H2O + Mg(NO3)2

593.2 grams Mg(NO3)2

Explanation:

First, balance the equation

The first thing I notice is that there are 2 nitrates on the reactant side so I would put a 2 in front of HNO3. Now the problem is that there are 4 hydrogens on the reactants side, so add a 2 to H2O.

Next is stoichiometry

We are looking for Mg(NO3)2 so use the balanced equation to create a mole ratio and then convert to grams.

8 mol HNO3 * 1 mol Mg(NO3)2/2 mol HNO3 * 148.3 g Mg(NO3)2/1 mol Mg(NO3)2 = 593.2 grams Mg(NO3)2

a). in this case, dh> 0 and sd> 0. if t becomes larger, the term "-tds" will become more negative. at some point, it is possible that -dts will be more negative than the dh is positive, and the reaction will become spontaneous. this is probably the correct answer, but you should look at all of the other possibilities and make sure of that.