Question

For the following reaction, 0.310 grams of hydrogen gas are allowed to react with 43.4 grams of iodine. hydrogen (g) + iodine (s) hydrogen iodide (g) What is the maximum amount of hydrogen iodide that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams

Answer 1

1)

a)

Molar mass of H2 = 2.016 g/mol

mass(H2)= 0.31 g

use:

number of mol of H2,

n = mass of H2/molar mass of H2

=(0.31 g)/(2.016 g/mol)

= 0.1538 mol

Molar mass of I2 = 253.8 g/mol

mass(I2)= 43.4 g

use:

number of mol of I2,

n = mass of I2/molar mass of I2

=(43.4 g)/(2.538*10^2 g/mol)

= 0.171 mol

Balanced chemical equation is:

H2 + I2 ---> 2 HI

1 mol of H2 reacts with 1 mol of I2

for 0.1538 mol of H2, 0.1538 mol of I2 is required

But we have 0.171 mol of I2

so, H2 is limiting reagent

we will use H2 in further calculation

Molar mass of HI,

MM = 1*MM(H) + 1*MM(I)

= 1*1.008 + 1*126.9

= 127.908 g/mol

According to balanced equation

mol of HI formed = (2/1)* moles of H2

= (2/1)*0.1538

= 0.3075 mol

use:

mass of HI = number of mol * molar mass

= 0.3075*1.279*10^2

= 39.34 g

Answer: 39.3 g

b)

H2 is limiting reagent

Answer: H2

C)

According to balanced equation

mol of I2 reacted = (1/1)* moles of H2

= (1/1)*0.1538

= 0.1538 mol

mol of I2 remaining = mol initially present - mol reacted

mol of I2 remaining = 0.171 - 0.1538

mol of I2 remaining = 1.723*10^-2 mol

Molar mass of I2 = 253.8 g/mol

use:

mass of I2,

m = number of mol * molar mass

= 1.723*10^-2 mol * 2.538*10^2 g/mol

= 4.373 g

Answer: 4.37 g

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