Chemical Reaction Calculation: Copper and Nitric Acid

How many grams of NO can be produced from the reaction of 3.18 g Cu and 15.75 g of HNO3 in the reaction below?

3 Cu + 8 HNO3 + 3 Cu(NO3)2 + 2 NO + 4 H2O

(A) 1.50 g (B) 1,00 g (C) 1.888 (D) 2.81 g (E)

Answer:

The reaction of 3.18 g Cu and 15.75 g HNO3 produces approximately 0.99 g of NO.

Explanation:

To calculate the grams of NO produced, we need to use the balanced equation and the given amounts of Cu and HNO3.

First, calculate the moles of Cu and HNO3 using their molar masses:

Molar mass of Cu = 63.55 g/mol

Molar mass of HNO3 = 63.01 g/mol

Moles of Cu = mass of Cu / molar mass of Cu = 3.18 g / 63.55 g/mol = 0.05 mol

Moles of HNO3 = mass of HNO3 / molar mass of HNO3 = 15.75 g / 63.01 g/mol = 0.25 mol

Next, use the mole ratio from the balanced equation to determine the moles of NO produced:

Mole ratio of Cu to NO = 3:2

Moles of NO = moles of Cu × (2 moles of NO / 3 moles of Cu) = 0.05 mol × (2/3) = 0.033 mol

Finally, calculate the grams of NO produced using the molar mass of NO:

Molar mass of NO = 30.01 g/mol

Grams of NO = moles of NO × molar mass of NO = 0.033 mol × 30.01 g/mol = 0.99 g

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