Calculate the Standard Enthalpy of the reaction below: NH3(g) + HCl (g) → NH4Cl(s)
Using the following Enthalpy of Reactions:
2HCl(g) → H2(g) + Cl2(g)
AH = +184.6 kJ
2H2(g) + 12 N2(g) + 2 Cl2(g) NH4Cl(s) AH = -314.4 kJ
N2(g) + 3 H2(g) → 2 NH3(g)
AH = +184.6 kJ​

(we use hess's law) it is so simple but the second reaction is not correct please right it

jdjxnbfb

explanation:

jjfnc fdnxnxx

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g