Question

For the following reaction, 15.1 grams of nitrogen monoxide are allowed to react with 5.41 grams of oxygen gas .

nitrogen monoxide(g) + oxygen(g) nitrogen dioxide(g)

What is the maximum mass of nitrogen dioxide that can be formed? ( grams )

What is the FORMULA for the limiting reagent?  

What mass of the excess reagent remains after the reaction is complete? (grams)

Answer 1

The given reaction is-

2NO (g) + O​​​​​​2 (g) -------> 2 NO​​​2 (g)

2 mols of nitrogen monoxide react with 1 mol of oxygen gas to form 2 mols of nitrogen di oxide.

#Molecular mass of-

• NO = 14+16 = 30 amu
• O​​​​​​2 = 16 × 2 = 32 amu
• NO​​​2 = 14 + (2×16) = 46 amu

Moles of any substance

= given mass (in gram) ÷ molecular mass

• Moles of NO = 15.1 ÷ 30 = 0.503 mols.
• Moles of O​​​​​​2 = 5.41 ÷ 32 = 0.169 mols.

Oxygen has lesser moles than nitrogen monoxide, so Oxygen gas is limiting reagent. It will decide the amount of nitrogen dioxide formed.

1 mol Oxygen gas produce nitrogen dioxide = 2 moles

therefore 0.169 mols Oxygen gas will produce = 2 × 0.169 = 0.338 mols of nitrogen dioxide.

mass of nitrogen dioxide = moles × molecular mass

= 0.338 × 46

= 15.548 grams nitrogen dioxide (maximum amount can be formed).

The formula of limiting reagent = O​​​​​​2

• Moles of excess reagent (nitrogen monoxide) remained = 0.503 - (2 × 0.169)

​​​​   = 0.165 moles.

• Mass of NO remaining (in grams)

= Moles × molecular mass

= 0.165 × 30 = 4.95 gram.

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