Thirty grams of lead oxide and fifteen grams of ammonia react completely to produce solid lead, nitrogen gas, and liquid water. What will be approximate mass of all three products?

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The balanced chemical equation for this reaction is given as:

`3 PbO + 2 NH_3 -> 3 Pb + N_2 + 3 H_2O`

Thus, 3 moles of lead oxide react with 2 moles of ammonia to generate 3 moles of lead, 1 mole of nitrogen gas and 3 moles of water. 

Molar mass of lead oxide = 207.2 + 16 = 223.2 gm/mole

Molar mass of ammonia = 14 + 3 x 1 = 17 gm/mole

Molar mass of lead = 207.2 gm/mole

Molar mass of nitrogen gas = 28 gm/mole

Molar mass of water = 18 gm/mole

Given moles of lead oxide = 30/223.2 = 0.1344 moles

Moles of ammonia = 15/17 moles = 0.8824 moles.

Since, 3 moles of PbO reacts with 2 moles ammonia, 0.1344 moles will react with 2/3 x 0.1344 = 0.0896 moles or say 0.09 moles.

Thus, PbO is in limiting quantity.

Using the given chemical reaction:

3 moles of PbO generate 3 moles of lead. Thus, 0.1344 moles of Pb or 27.85 gm (= 0.1344 x 207.2) Pb will be produced.

Similarly, 3 moles of PbO produce 1 mole of nitrogen gas. Thus, 0.0448 moles (= 1/3 x 0.1344) or 1.2544 gm (or 1.25 gm) nitrogen gas will be generated.

Similarly, 0.1344 moles or 2.42 gm (=0.1344 x 18) water will be generated.

Hope this helps.

Approved by eNotes Editorial Team

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