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Dominik [7]
3 years ago
10

How many molecules of carbon monoxide (co) are needed to react with excess iron (iii) oxide (fe2o3) to produce 11.6 g of iron (f

e)?
Chemistry
1 answer:
Troyanec [42]3 years ago
7 0
<span>8.73 grams CO
   The balanced equation for the reaction is
3CO + Fe2O3 ==> 2Fe + 3CO2
   So for every 2 atoms of Fe produced, 3 molecules of CO is needed. So let's start by looking up the atomic weights of the involved elements:
 Atomic weight iron = 55.845
  Atomic weight carbon = 12.0107 
  Atomic weight oxygen = 15.999
   Molar mass CO = 12.0107 + 15.999 = 28.0097 g/mol
 Moles iron = 11.6 g / 55.845 g/mol = 0.20771779 mol
   Moles CO needed = 0.20771779 mol / 2 * 3 = 0.311576685 mol
 Mass CO needed = 0.311576685 mol * 28.0097 g/mol = 8.727169487 g Rounding to 3 significant figures gives 8.73 grams CO</span>
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A compound of mercury and oxygen is heated in order to decompose the compound. A 4.08 grams sample of mercury oxide upon heating
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<span>Hope this answers your question!</span>
4 0
3 years ago
How many grams of sulfuric acid is needed to neutralize 380 ml of solution with pH = 8.94
erma4kov [3.2K]

Answer : The mass of sulfuric acid needed is 16.23\times 10^{-5}g.

Solution : Given,

pH = 8.94

Volume of solution = 380 ml = 380\times 10^{-3}      (1ml=10^{-3}L)

Molar mass of sulfuric acid = 98.079 g/mole

As we know,

pH+pOH=14\\pOH=14-8.94=5.06

pOH=-log[OH^-]

5.06=-log[OH^-]

[OH^-]=0.00000871=8.71\times 10^{-6}mole/L

Now we have to calculate the moles of OH^-.

Formula used : Moles=Concentration\times Volume

\text{ Moles of }[OH^-]=\text{ Concentration of }[OH^-]\times Volume\\\text{ Moles of }[OH^-]=(8.71\times 10^{-6}mole/L)\times (380\times 10^{-3}L)=3309.8\times 10^{-9}moles

For neutralization, equal number of moles of H^+ ions will neutralize same number of OH^- ions.

\text{ Moles of }[OH^-]=\text{ Moles of }[H^+]=3309.8\times 10^{-9}moles

As, H_2SO_4\rightarrow 2H^++SO^{2-}_4

From this reaction, we conclude that

2 moles of H^+ ion is given by the 1 mole of H_2SO_4

3309.8\times 10^{-9} moles of H^+ ion is given by \frac{3309.8\times 10^{-9}}{2}=1654.9\times 10^{-9} moles of H_2SO_4

Now we have to calculate the mass of sulfuric acid.

Mass of sulfuric acid = Moles of H_2SO_4 × Molar mass of sulfuric acid

Mass of sulfuric acid = (1654.9\times 10^{-9}moles)\times (98.079g/mole)=162310.94\times 10^{-9}=16.23\times 10^{-5}g

Therefore, the mass of sulfuric acid needed is 16.23\times 10^{-5}g.

3 0
3 years ago
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