Question

12. Potassium reacts aggressively with water, as shown in this equation: 2 K + 2 H20→2 KOH + H2 If 0.2 grams of potassium are
13. A chemist carries out the following reaction: C.Hg + Br2 → C.HsBr + HBr a. If he reacts 14.2 grams of CH with an excess o
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Answer #1

12)

Molar mass of K = 39.1 g/mol

mass(K)= 0.2 g

use:

number of mol of K,

n = mass of K/molar mass of K

=(0.2 g)/(39.1 g/mol)

= 5.115*10^-3 mol

Molar mass of H2O,

MM = 2*MM(H) + 1*MM(O)

= 2*1.008 + 1*16.0

= 18.016 g/mol

mass(H2O)= 15.0 g

use:

number of mol of H2O,

n = mass of H2O/molar mass of H2O

=(15 g)/(18.02 g/mol)

= 0.8326 mol

Balanced chemical equation is:

2 K + H2O ---> 2 KOH + H2

2 mol of K reacts with 1 mol of H2O

for 5.115*10^-3 mol of K, 2.558*10^-3 mol of H2O is required

But we have 0.8326 mol of H2O

so, K is limiting reagent

we will use K in further calculation

Molar mass of KOH,

MM = 1*MM(K) + 1*MM(O) + 1*MM(H)

= 1*39.1 + 1*16.0 + 1*1.008

= 56.108 g/mol

According to balanced equation

mol of KOH formed = (2/2)* moles of K

= (2/2)*5.115*10^-3

= 5.115*10^-3 mol

use:

mass of KOH = number of mol * molar mass

= 5.115*10^-3*56.11

= 0.287 g

Answer: 0.287 g

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