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)
-
1.12dm3
- 2.24dm3
- 4.48dm3
- 11.2dm3
- 22.4dm3
The correct answer is: A
Explanation
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\(_2\)