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viva [34]
3 years ago
10

How many moles of aluminum are needed to react completely with 1.2 mol of feo

Chemistry
1 answer:
Amanda [17]3 years ago
3 0

Answer:

0.8 mol.

Explanation:

  • The balanced equation for the reaction between Al and FeO is represented as:

<em>2Al + 3FeO → 3Fe + Al₂O₃,</em>

It is clear that 2 mol of Al react with 3 mol of FeO to produce 3 mol of Fe and 1 mol of Al₂O₃.

<em><u>Using cross multiplication:</u></em>

2 mol of Al needs → 3 mol of FeO, from stichiometry.

??? mol of Al needs → 1.2 mol of FeO.

∴<em> The no. of moles of Al are needed to react completely with 1.2 mol of FeO </em>= (2 mol)(1.2 mol)/(3 mol) = <em>0.8 mol.</em>

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The density of no2 in a 4.50 l tank at 760.0 torr and 25.0 °c is ________ g/l.
kap26 [50]
From  ideal  gas  equation   that   is   PV=nRT
n(number of  moles)=PV/RT
P=760 torr
V=4.50L
R(gas  constant =62.363667torr/l/mol
T=273 +273=298k
n  is   therefore   (760torr x4.50L) /62.36367 torr/L/mol  x298k  =0.184moles
the  molar  mass  of  NO2 is  46  therefore  density=  0.184  x  46=8.464g/l
7 0
3 years ago
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nikitadnepr [17]

Answer:

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Explanation:

7 0
3 years ago
. What mass of ammonium chloride must be added to 250. mL of water to give a solution with pH
Eduardwww [97]

The mass of ammonium chloride that must be added is : ( A ) 4.7 g

<u>Given data :</u>

Volume of water ( V )  = 250 mL = 0.25 L

pH of solution = 4.85

Kb = 1.8 * 10⁻⁵

Kw = 10⁻¹⁴

Given that the dissolution of NH₄Cl gives NH₄⁺⁺ and Cl⁻ ions the equation is written as :

NH₄CI  +  H₂O  ⇄  NH₃ + H₃O⁺

where conc of H₃O⁺

[ H₃O⁺ ] = \sqrt{Ka.C}   and Ka = Kw / Kb

∴ Ka = 5.56 * 10⁻¹⁰

Next step : Determine the concentration of H₃O⁺  in the solution

pH = - log [ H₃O⁺ ] = 4.85

∴ [ H₃O⁺ ] in the solution = 1.14125 * 10⁻⁵

Next step : Determine the concentration of NH₄CI in the solution

C = [ H₃O⁺ ]² / Ka

  = ( 1.14125 * 10⁻⁵ )² /  5.56 * 10⁻¹⁰

  = 0.359 mol / L

Determine the number of moles of NH₄CI in the solution

n = C . V

  = 0.359 mol / L  * 0.25 L =  0.08979 mole

Final step : determine the mass of ammonium chloride that must be added to 250 mL

mass = n * molar mass

         = 0.08979 * 53.5 g/mol

         = 4.80 g  ≈ 4.7 grams

Therefore we can conclude that the mass of ammonium chloride that must be added is 4.7 g

Learn more about ammonium chloride : brainly.com/question/13050932

8 0
2 years ago
The decomposition of ethyl amine, C2H5NH2, occurs according to the reaction: C2H5NH2(g)⟶C2H4(g)+NH3(g) At 85∘C, the rate constan
statuscvo [17]

Answer:

2.772 seconds

Explanation:

Given that;

t1/2 = 0.693/k

Where;

t1/2 = half life of the reaction

k= rate constant

Note that decomposition is a first order reaction since the rate of reaction depends on the concentration of one reactant

t1/2 = 0.693/2.5 x 10-1 s-1

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7 0
3 years ago
A chemist makes 940. mL of sodium carbonate (Na_2CO_3) working solution by adding distilled water to 200. mL of a 0.171 m stock
Alexxandr [17]

Answer:

The answer is 0.0698 M

Explanation:

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The formula for the preparation I M1V1 = M2V2

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V1= volume of the stock solution taken = 200 mL

M2 = the concentration produced

V2 = the volume of the solution produced = 940 mL

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The concentration of the Chemist's working solution to 3 significant figures is 0.0698M

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