Answer to Question #196997 in General Chemistry for lea

Question

If the percent yield for the following reaction is 65.0%, how many grams of KClO₃ are needed to produce 6.00 g of O₂?

2 KClO₃(s) → 2 KCl(s) + 3 O₂(g)

a. 24.0 g

b. 9.95 g

c. 52.9 g

d. 15.3 g

e. 23.6 g

Magnesium hydroxide, the active ingredient in milk of magnesia, neutralizes stomach acid, primarily HCl, according to the reaction: Mg(OH)2(aq) + 2 HCl(aq) → H2O(l) + MgCl2(aq) How much HCl in grams can be neutralized by 3.43 g of Mg(OH)2 ?

mass = ----g

Answer 1

2 KClO₃(s) → 2 KCl(s) + 3 O₂(g)

Mol mass of KClO₃ = 39+35.5+3×16=39+35.5+48 =122.5

3x16x2 g of oxygen is produced from 2x122.5 KClO₃

Or 96g of Oxygen produce 245 of KClO₃

1 g of Oxygen is produced from

$\frac{245}{96}×6$

15.31g of KClO₃ is required if the yield is 100%

since percent yield is 65.0 %

therefore, actual yield will be $\frac{15.31}{100}×65=9.95g$

Hence, option b is correct

 Mg(OH)_2(aq) + 2 HCl_(aq) → H₂O_(l) + MgCl_2(aq)

Mol mass of Mg(OH)₂=24+2(16+1)=24+34=58

HCl = 35.5+1=36.5

58 g of Mg(OH)₂ neutralises 36.5g of HCl

1 g of Mg(OH)₂neutralise =$\frac{36.5}{58}$

3.43g of Mg(OH)₂ =$\frac{36.5}{58}×3.43$

2.15g of HCl is produced