The reaction of silver metal and dilute nitric acid proceeds according to the equation above. if 0.10 mole of powdered silver is added to 10.0 milliliters of 6.0 molar nitric acid, the number of moles of no gas that can be formed is

a) 0.015 mole

b) 0.020 mole

c) 0.030 mole

d) 0.045 mole

e) 0.090 mole

The moles of NO formed from the given reaction has been 0.015 mol. Thus, option A is correct.

The balanced chemical equation for the reaction has been:

The moles of nitric acid can be given as:

Moles = Molarity × Volume

The molarity of nitric acid = 6 M

Volume of nitric acid = 10 ml = 0.01 L

Moles of nitric acid = 6 × 0.01 mol

Moles of nitric acid = 0.06 mol.

From the balanced equation,

3 moles silver requires = 4 moles nitric acid

0.10 moles silver requires = moles nitric acid

0.10 moles silver requires = 0.13 mol nitric acid.

Since the available nitric acid is 0.06 mol, it has been in lesser amount and has been the limiting reagent.

Thus,

4 moles nitric acid = 1 moles NO

0.06 mol nitric acid = 0.015 moles NO.

Thus, the moles of NO formed from the given reaction has been 0.015 mol. Thus, option A is correct.

For more information about the moles in a equation, refer to the link:

link

The answer to your question is: 0.015 mole of NO

Explanation:

Data

Silver = Ag = 0.1 mole

Nitric acid = HNO3 = 10 ml   6 M

Balanced reaction

                     3Ag    +  4HNO₃    ⇒     3AgNO₃   +    NO    +     2H₂O

# moles of HNO₃ = 6(0.01) = 0.06

Limiting reactant   HNO₃  because for the amount of silver given

                              the reaction  needs 0.133 moles of acid, and  there

                             are only 0.06

                    4 mol of HNO₃    1 mol of NO

                   0.06 mol of Ag      x

                   x = (0.06 x 1) / 4

                   x = 0.015 mol

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