Answer to Question #196997 in General Chemistry for lea

Question #196997

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

Expert's answer

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


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}\u00d765=9.95g"

Hence, option b is correct

 Mg(OH)2(aq) + 2 HCl(aq) → H2O(l) + MgCl2(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}\u00d73.43"

2.15g of HCl is produced

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