For the following reaction, 13.6 grams of chlorine gas are allowed to react with

Question

For the following reaction, 13.6 grams of chlorine gas are allowed to react with 62.3 grams of sodium iodide chlorine(g) + sodium iodide(s)sodium chloride(s) + iodine(s) What is the maximum amount of sodium chloride 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 An error has been detected in your answer. Check for typos, miscalculations etc. before submitting your answer. Submit Answer Retry Entire Group No more group attempts remain

Explanation / Answer

Molar mass of Cl2 = 70.9 g/mol


mass(Cl2)= 13.6 g

use:
number of mol of Cl2,
n = mass of Cl2/molar mass of Cl2
=(13.6 g)/(70.9 g/mol)
= 0.1918 mol

Molar mass of NaI,
MM = 1*MM(Na) + 1*MM(I)
= 1*22.99 + 1*126.9
= 149.89 g/mol


mass(NaI)= 62.3 g

use:
number of mol of NaI,
n = mass of NaI/molar mass of NaI
=(62.3 g)/(149.89 g/mol)
= 0.4156 mol
Balanced chemical equation is:
Cl2(g) + 2 NaI(s) —> 2 NaCl(s) + I2(s)


1 mol of Cl2 reacts with 2 mol of NaI
for 0.1918 mol of Cl2, 0.3836 mol of NaI is required
But we have 0.4156 mol of NaI

so, Cl2 is limiting reagent
we will use Cl2 in further calculation


Molar mass of NaCl,
MM = 1*MM(Na) + 1*MM(Cl)
= 1*22.99 + 1*35.45
= 58.44 g/mol

According to balanced equation
mol of NaCl formed = (2/1)* moles of Cl2
= (2/1)*0.1918
= 0.3836 mol


use:
mass of NaCl = number of mol * molar mass
= 0.3836*58.44
= 22.4 g

According to balanced equation
mol of NaI reacted = (2/1)* moles of Cl2
= (2/1)*0.1918
= 0.3836 mol
mol of NaI remaining = mol initially present - mol reacted
mol of NaI remaining = 0.4156 - 0.3836
mol of NaI remaining = 3.2*10^-2 mol


Molar mass of NaI,
MM = 1*MM(Na) + 1*MM(I)
= 1*22.99 + 1*126.9
= 149.89 g/mol

use:
mass of NaI,
m = number of mol * molar mass
= 3.2*10^-2 mol * 149.89 g/mol
= 4.80 g
Answers:
1. 22.4 g
2. Cl2
3. 4.80 g

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