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 added to 15.0 grams of water, what mass of KOH can form in this reaction? 7.4 Theoretical and Percent Yield 13. A chemist carries out the following reaction: C.He + Brı → C.H.Br + HBr a. If he reacts 14.2 grams of GoHs with an excess of Brz, what is the theoretical yield of C.H.Br? b. If he isolated 16.3 grams of CoHsBr, what is his percent yield for this reaction?

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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