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Flauer [41]
3 years ago
9

Nitric acid reacts with silver metal 4hno3 + 3ag = no + 2h2o + 3agno3. calculate the number of grams of no formed when 10.8 g of

ag reacts with 12.6g of hno3
Chemistry
1 answer:
blagie [28]3 years ago
3 0
The balanced equation for the reaction is;
<span>2HNO</span>₃<span> + Ag ---> NO + 2H</span>₂O<span> + AgNO</span>₃
Stoichiometry of HNO₃ to Ag is 2:1
we need to first find the limiting reactant
number of HNO₃ moles - 12.6 g / 63 g/mol  = 0.2 mol
number of Ag moles - 10.8 g / 108 g/mol = 0.1 mol
0.2 moles of HNO₃ react with 0.1 mol of Ag, this is in the 2:1 molar ratio which means that both reactants react fully in the reaction.
then molar ratio of Ag to NO is 1:1
number of Ag moles equivalent to number of NO moles 
Number of NO moles - 0.1 mol
Mass of NO - 0.1 mol x 30 g/mol = 3 g
mass of NO produced is 3 g


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Answer:

Therefore the concentration of the reactant after 4.00 minutes will be 0.686M.

Explanation:

The unit of k is s⁻¹.

The order of the reaction = first order.

First order reaction: A first order reaction is  a reaction in which the rate of reaction depends only the value of the concentration of the reactant.

-\frac{d[A]}{dt} =kt

[A] = the concentration of the reactant at time t

k= rate constant

t= time

Here k= 4.70×10⁻³ s⁻¹

t= 4.00

[A₀] = initial concentration of reactant = 0.700 M

-\frac{d[A]}{dt} =kt

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Integrating both sides

\Rightarrow\int -\frac{d[A]}{[A]}=\int kdt

⇒ -ln[A] = kt +c

When t=0 , [A] =[A₀]

-ln[A₀]  = k.0 + c

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Therefore

-ln[A] = kt - ln[A₀]

Putting the value of k, [A₀] and t

- ln[A] =4.70×10⁻³×4 -ln (0.70)

⇒-ln[A]=  0.375

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5 0
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Read 2 more answers
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Evgesh-ka [11]

Answer:

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Explanation:

Step 1: Given data

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Step 2: Calculate the volume occupied by aluminum

The density of aluminum is equal to its mass divided by its volume.

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V = 30 cm³

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