Read the given equation.

na2o2 + co2 → na2co3 + o2

what volume of o2 gas is produced from 4.60 liters of co2, assuming all values are measured at stp?

1.20 liters
2.30 liters
5.20 liters
9.20 liters

2.30 liters.

Explanation:

Na₂O₂ + CO₂ → Na₂CO₃ + 1/2O₂,

P is the pressure of the gas (P = 1.0 atm at STP),

V is the volume of the gas (V = 4.60 L),

R is the general gas constant (R = 0.082 L.atm/mol.K),

T is the temperature of the gas (T = 273.0 K at STP).

∴ n = PV/RT = (1.0 atm)(4.6 L) / (0.082 L.atm/mol.K)(273.0 K) = 0.205 mol.

Using cross multiplication:

1.0 mole of CO₂ produces → 0.5 mole of O₂, from the stichiometry.

0.205 mole of CO₂ produces → ??? mole of O₂.

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

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