Question

Use the equation 6KI + SHNO3 ----> 6KNO3 + 2NO + 3I2 + 4H2O. Note that 1 mole of any gas is measured at STP will occupy 22.4L.

a.) If 38g of KI are reacted, what mass of KNO3 will form?

b.) What volume of NO gas, measured at STP, will produced if 37.0g of HNO3 are consumed?

c.) If 0.50 mole of KI is to be reacted, what volume (in mL) of 6.00 M HNO3 will be required?

d.) When the reaction produces 6.0 moles of NO, how many molecules of I2 will be produced?

e.) How many grams of molecular iodine can be obtained by reacting 45.0mL of a 0.350 M KI solution?

Answer 1

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

a)
according to balanced equation 6 mol KI produce 6 mol KNO3 [1:1 mol ratio]

mols of KI in 38 g KI = 38 g/166 g/mol = 0.2289 mols

mass of KNO3 in 0.2289 mol =0.2289 mol *101.10 g/mol = 23.1 g
Answer
23.1 g
***********
b)
mols of HNO3 in 37 g= 37 g/63.01 g/mol = 0.5872 mols

mols of NO produced = 1/4 * 0.5872 mol = 0.1468 mol
volume of NO = 0.1468 mol * 22.4 L
= 3.29 mol
************************
c) from balanced equation 6 mol KI need 8 mol HNO3
0.5 mol KI need 8 mol/6mol * 0.5 mol = 0.666 mol
volume of HNO3 needed = 0.666 mol/6 M = 0.111 L
volume needed = 111 ml
****************
d)
mol ratio NO : I2 = 2:3
mols of I2 produced from 6 mols of NO = 3/2 * 6 mol = 9 mol

Number of I2 molecules = 6.02 x10^23 * 9 mol
= 54.2 x10^23 molecules
*********************************
e)
Number of mols of KI used = 0.045 L * 0.35 M
= 0.01575 mols
mols of I2 formed = 3/6 * 0.01575 mols
= 0.007875 mols
mass of I2 formed = 0.007875 mols *253.81 g/mol
= 1.998 g

mass of I2 formed = 2.00 g
************************
Thanks!