2NaNO3 → 2NaNO2 + O2. From the equation above, what is the volume of oxygen produced at s.t.p. when 8.5g of 2NaNO3 is heated until no further gas is evolved? (2NaNO3 = 85, molar volume of gases at s.t.p. = 22.4dm3)

  • A 1.12dm3
  • B 2.24dm3
  • C 4.48dm3
  • D 11.2dm3
  • E 22.4dm3
waec 1991chemistry

The correct answer is A. 1.12dm3

85g of 2NaNO\(_3\) will liberated 22.4dm\(_3\)

8.5g = xdm\(_3\)

cross multiply: x = [8.5 * 22.4] / 85

x = 2.24dm\(_3\)

: 8.5g of  2NaNO\(_3\) = 2.24dm\(_3\)

where 2moles of NaNO\(_3\) liberates 1mole of O\(_2\)

2.24dm\(_3\) of NaNO\(_3\) liberates 1.12dm\(_3\) of O\(_2\)

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