JEE Main 2022 · 29 Jul · Shift 2 · Chemistry Q61

ChemistrySingle correct+4 / -1Question 61 of 90

Consider the reaction

$\mathrm{4HNO_3(l)+3KCl(s)\to Cl_2(g)+NOCl(g)+2H_2O(g)+3KNO_3(s)}$

The amount of HNO $_3$ required to produce 110.0 g of KNO $_3$ is

(Given: Atomic masses of H, O, N and K are 1, 16, 14 and 39, respectively.)

32.2 g

69.4 g

91.5 g

162.5 g

Answer and worked solution

Official answer: C

Moles. KNO₃: $\frac{110}{101}=1.089$ mol, so HNO₃ $=\frac43\times1.089=1.452$ mol $=91.5$ g.

(C)

Topics:Stoichiometry

JEE Main 2022 · 29 Jul · Shift 2, Chemistry question 61