Percent yield of `NH_(3)` in the following reaction is 80%
`NH_(2)CONH_(2) + 2NaOH overset(Delta)rarrNa_(2)CO_(3)+2NH_(3)`
`6g NH_(2)CONH_(2)` reacts with 8g NaOH to form `NH_(3)`. Calculate mass of `NH_(3)` formed.

Options

A. 3.4 g
B. 2.72 g
C. 4.25 g
D. 11.2 g

Answer

B

Explanation

`NH_(2)CONH_(2)+2NaOHoverset(80%)rarrNa_(2)CO_(3)+2NH_(3)`
`0.1 " mol" " "0.2 " mol"`
`(n_("urea"))/(1)xx(80)/(100)=n_(NH_(3))/(2)`
`n_(NH_(3)) = 0.16`
`n_(NH_(3)) = 0.16xx17 = 2.72g`