What volume of NO gas (at STP) will be produced if 45.0g of HNO3 react with excess potassium iodide?

Question

8HNO3 + 6KI>>>3 I2 + 2NO+6KNO3+4H2O

Answer 1

MM( HNO3 ) = 63 g/mol

45.0 g / 63 = 0.714 mole HNO3

the ratio between HNO3 and NO is 8 : 2

8 : 2 = 0.714 : x

x = 0.178 mole NO produced

pV = nRT

at STP
p=1 atm
T = 273 K

V = nRT / p = 0.178 x 0.0821 x 273 / 1 = 4.00 L NO produced

Answer 2

45/63 x 2/8 x 22.4