Na₂CO₃ + 2HCI → 2NaCI + H₂O + CO₂. Using the above equation, what volume of carbondioxide, measured at s.t.p is liberated when 53g of sodium carbonate is dissolved in hydrochloric acid?
(1 mole of gas occupies 22.4 dm³ at s.t.p)
(Na = 23, C = 12, O = 16)

olyezema

4 years ago

n=m/M
M=23×2+12+48
n=53/106=0.5moles
Na2CO3+2HCl ------ 2NaCl+CO2+H2O
1mole ------- 22.4dm³
0.5mole --------- 0.5×22.4=11.2dm³

Jokesta

2 years ago

First, we need to calculate the number of moles of sodium carbonate:

m = 53g
M = (2x23) + 12 + (3x16) = 106g/mol
n = m/M = 53/106 = 0.5 moles

According to the balanced equation, 1 mole of sodium carbonate produces 1 mole of carbon dioxide. Therefore, 0.5 moles of sodium carbonate produces 0.5 moles of carbon dioxide.

Using the ideal gas law, we can calculate the volume of carbon dioxide at standard temperature and pressure (STP):
PV = nRT

where P = 1 atm, V = the unknown volume, n = 0.5 moles, R = 0.082 L·atm/K·mol (the gas constant), and T = 273 K (the temperature at STP).

Solving for V, we get:

V = (nRT)/P
V = (0.5)(0.082)(273)/1
V = 11.22 L

Therefore, 11.22 liters of carbon dioxide will be liberated when 53g of sodium carbonate is dissolved in hydrochloric acid.

odewenusofiyyah

2 years ago

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