Zinc metal reacts with hydrochloric acid to produce hydrogen gas. How many liters of

hydrogen can be produced at STP if 13.5g of zinc are reacted with excess HCI?

Zn + 2 HCI → H2 + ZnCl

4.65L

Explanation:

Equation of reaction,

Zn + 2HCl → H₂ + ZnCl₂

From the equation of reaction,

1 mole of Zn produce 1 mole of H₂

Number of mole = mass / molar mass

Molar mass of Zn = 65g/mol

Molar mass of H₂ = 2.0g/mol

Mass of Zn = 65g

Mass of H₂ = 2.0g

Therefore,

65g of Zn will produce 2g of H₂

13.5g of Zn will produce yg of H₂

Y = (13.5 × 2) / 65

Y = 0.415g of H₂

13.5g of Zn will produce 0.415g of H₂

From ideal gas equation,

PV = nRT

P = pressure of the gas

V = volume the gas occupies

n = number of moles of the gas

R = ideal gas constant = 0.082J/mol.K

T = temperature of the gas

At STP,

P = 1.0atm

V = ?

n = number of moles = mass / molar mass = 0.415 / 2 = 0.2075moles

R = 0.082J/mol.K

T = 273.15K

PV = nRT

V = nRT / P

V = (0.2075 × 0.082 × 273.15) / 1

V = 4.647L

Volume of the gas occupies is approximately 4.65L

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