In two or more complete sentences explain how to balance the chemical equation and classify its reaction type. ___H3PO4 + ___NaBr ⟶ ___HBr + ___Na3PO4

Double displacement reaction

Explanation:

Chemical equation:

H₃PO₄ + NaBr      →    HBr + Na₃PO₄

Balanced chemical equation:

H₃PO₄ + 3NaBr      →    3HBr + Na₃PO₄

Step one:

H₃PO₄ + NaBr      →    HBr + Na₃PO₄

On left side of equation                  On right side pf equation

H = 3                                                   H = 1

P =  1                                                    P = 1

O = 4                                                    O = 4

Na = 1                                                   Na = 3

Br = 1                                                     Br = 1

Step 2:

H₃PO₄ + NaBr      →    3HBr + Na₃PO₄

On left side of equation                  On right side pf equation

H = 3                                                   H = 3

P =  1                                                    P = 1

O = 4                                                    O = 4

Na = 1                                                   Na = 3

Br = 1                                                     Br = 3

Step 3:

H₃PO₄ + 3NaBr      →    3HBr + Na₃PO₄

On left side of equation                  On right side pf equation

H = 3                                                   H = 3

P =  1                                                    P = 1

O = 4                                                    O = 4

Na = 3                                                   Na = 3

Br = 3                                                     Br = 3

Type of chemical reaction:

The given reaction is double displacement reaction on which cation and anion of both reactants are replace with each other.

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