1. if we have a reaction of zn3po4 and hcl, what would the products of thisreaction be? if we produced 13.05 g of h3po4, how many grams ofhydrochloric acid would be need to start with?

1. The products of this reaction are ZnCl₂ and H₃PO₄.

2. 14.57 g.

Explanation:

1. What would the products of this  reaction be?

Zn₃(PO₄)₂ + 6HCl → 3ZnCl₂ + 2H₃PO₄,

It is clear that 1.0 mol of Zn₃(PO₄)₂ reacts with 6.0 mol of HCl to produce 3.0 mol of ZnCl₂ and 2.0 mol of H₃PO₄.

So, the products of this reaction are ZnCl₂ and H₃PO₄.

2. If we produced 13.05 g of H₃PO₄, how many grams of  hydrochloric acid would be need to start with?

n = mass/molar mass = (13.05 g)/(97.994 g/mol) = 0.1332 mol.

Using cross-multiplication:

6.0 mol of HCl needed to produce → 2.0 mol of H₃PO₄, from stichiometry.

??? mol of HCl needed to produce → 0.1332 mol of H₃PO₄.

∴ The no. of moles of HCl needed = (6.0 mol)(0.1332 mol)/(2.0 mol) = 0.3995 mol.

∴ The mass of HCl needed = n*molar mass = (0.3995 mol)(36.46 g/mol) = 14.57 g.

So, the grams of  hydrochloric acid would be need to start with = 14.57 g.

"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"

pie chart because it is in percents ( % )