Question

3 Ag(s) + 4 HNO3 → 3 AgNO3 + NO(g) + 2 H2OThe 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. milliliters of 6.0-molar nitric acid, the number of moles of NO gas that can be formed is

Answer 1

`0.015\text{ mol}`

Step-by-step explanation:

Here, we want to get the number of moles of NO gas that can be formed

To answer this, we need to be sure of the limiting reagent

The limiting reagent is the one that is responsible for the number of moles of product formed

From the shown balanced equation of reaction, 3 moles of silver produced 1 mole of the gas

Thus, 0.1 mol of powdered silver will produce:

`(0.1)/(3)\text{ = 0.033 mol}`

For the nitric acid, we need to get the number of moles that reacted

To get this, we have to multiply the volume (in L) that reacted by the molarity

Mathematically, we have that as:

`\begin{gathered} 10\text{ ml = }(10)/(1000)\text{ = 0.01 L} \\ \\ number\text{ of moles = volume }*\text{ molarity} \\ number\text{ of moles = 0.01 }*\text{ 6 = 0.06 moles} \end{gathered}`

From the equation of reaction, 4 moles of the nitrate gave 1 mole of the gas

Thus, we have it that 0.06 moles will give:

`(0.06)/(4)\text{ = 0.015 mole}`

From what we see, the number of moles produced by the nitrate is less than what was produced by the solid

Thus, the nitrate is the limiting reagent and the number of moles of the gas produced is 0.015 mol