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1How many milligrams of water are produced when 300 millimeters 0.2M sulfuric ac

ID: 899509 • Letter: 1

Question

1How many milligrams of water are produced when 300 millimeters 0.2M sulfuric acid are mixed with one gallon of 0.002 M Calcium hydroxide .The reaction took place with 25% of efficiency. The reaction products are water and Calcium sulfate.

2Why are atomic weights relative weights?

3 How many grams of Fe were required if a reaction of Iron and enough HCl produced 0.42lbs. of FeCl2. The reaction took place with 37% of efficiency. The equation is not balanced.1kg = 2.2lbs

Fe(s) + HCl (aq) CaCl2 (aq) + H2 (g)

Explanation / Answer

NOTE that 300 milimeters are probably mililiters so...

V = 300 ml

M = 0.2 M of H2SO4

V = 1 gal or 3.79 l

M = 0.002 M of Ca(OH)2

25% conversion

H2SO4 + Ca(OH)2 ---> 2H2O + CaSO4(aq)

calculate moles of H2O produced:

mol of acid = 300*.2 = 60 mmol of acid

mol of base = 3.79*.02*1000 = 75.8 mmol of base

ratio is 1:1 so acid is limiting reactant

60 mmol of acid react to form 2*60 mmol of h2o; 120 mmol of water will be produced

120 mmol --> 0.12 mol ofH2O

But on ly 25% is actually produced so 0.12*0.25 = 0.03

MW of H2O = 18

mass = mol*MW = 0.03*18 = 0.54 grams of H2O, if you need mg, x1000, 540 mg of H2O produced

2Why are atomic weights relative weights?

they are relative since they are related to their mole:mass ratio i.e.

if you have Carbon MW = 12

this mean this is 12 grams per gmol, 12 lb per lbmol, 12 kg per kmol, etc...

They are related to their ratios

3)

Fe(s) + 2HCl (aq) FeCl2 (aq) + H2 (g)

m = Fe is

mass of FeCl2 = 0.42 lb or 0.191 kg or 191 grams of FeCl2 are produced

n = 37%

Note that if we porduced only 37% then

191 grams, fi reacted completely, should be 191/0.37 = 516.2 grams in total

do the balance based on that

MW of FeCl2 = 126.751g/mol

mol in 516.2 grams --> 516.2/126.751 = 4.07 mol of FeCl2 produced

since ratio is 1:1 (1 mol of Fe to produce 1 mol of FeCl2)

then, we had 4.07 mol of FE,

MW of Fe = 55 g/mol

mass = mol*MW = 4.07*55 = 223.85 grams of Fe are needed

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