10.65 g of Cl2 reacts with excess ammonia according to 8NH3 + 3Cl2 -> 6NH4Cl + N2. Taking N=14, H=1 and Cl=35.5, the mass of NH4Cl formed is

JEE Advanced levelSingle correct optionDifficulty 3 of 3ph105

$10.65 g$ of $Cl X_{2}$ reacts with excess ammonia according to $8 NH X_{3} + 3 Cl X_{2} 6 NH X_{4} Cl + N X_{2}$ . Taking $N = 14$ , $H = 1$ and $Cl = 35.5$ , the mass of $NH X_{4} Cl$ formed is

Moles of

Cl X_{2} = 10.65/71 = 0.15

. The equation gives 6 mol

NH X_{4} Cl

per 3 mol

Cl X_{2}

, i.e. 2 mol per mol, so

0.30

mol

NH X_{4} Cl

. Molar mass is

53.5

, so mass

= 16.05 g