Question

1. [12.7 g C₄H₁₀] Butane, C₄H₁₀, is a common fuel. How many grams of butane can be burned by 45.4 grams of oxygen?

            2 C₄H₁₀ + 13 O₂ ---> 8 CO₂ + 10 H₂O

2. [350. g NH₄NO₃] The fertilizer ammonium nitrate (NH₄NO₃) can be made by direct combination of ammonia with nitric acid:

            NH₃ + HNO₃ ---->  NH₄NO₃

If 74.4 grams of ammonia (NH₃) is reacted with nitric acid, how many grams of ammonium nitrate can be produced?

3. [710. g NaHCO₃] What mass of NaHCO₃ must decompose to produce 448 grams of Na₂CO₃?

            2 NaHCO₃ ---->  Na₂CO₃ + H₂O + CO₂

Answer 1

1) 2C4H10 + 13O2 -----> 8CO2 + 10H2O

Here we can see from 13 moles of oxygen, 2 moles of butane is burned.

So from 1 mole of oxygen number of moles of butane will be burned= (2/13) moles

Here mass of oxygen = 45.4 grams

Molar mass of oxygen= 32 g/mole

So the number of moles of oxygen present= 1.419 moles

So the number of moles of butane will be burned= [1.419×(2/13)] moles = 0.218 moles

Molar mass of butane = 58.12 g/mole

So the mass of butane can be burned with 45.4 grams of oxygen= (0.218×58.12) grams = 12.7 grams. (answer)

2) NH3 + HNO3 = NH4NO3

Here from 1 mole of ammonia, one mole of NH4NO3 is produced.

Mass of ammonia= 74.4 g

Molar mass of ammonia= 17.031g/mole

So number of moles of ammonia present= (74.4/17.031) moles = 4.37 moles

So NH4NO3 will be produced= 4.37 moles

Molar mass of NH4NO3 = 80.043 g/mole

So number of grams of ammonium nitrate(NH4NO3) will be produced= (4.37×80.043) grams = 350 grams(answer)

3) 2NaHCO3 -----> Na2CO3 + H2O +CO2

To produce 1 mole of Na2CO3 2 moles of NaHCO3 is required.

Here mass of Na2CO3 = 448 grams

Molar mass of Na2CO3 = 105.99 g/mole

So number of moles of Na2CO3 =(448/105.99) moles = 4.23 moles

So NaHCO3 required to produce 4.23 moles of Na2CO3 = (4.23×2)moles = 8.46 moles

Now molar mass of NaHCO3 = 84.007 g/mole

So mass of NaHCO3 required to decompose to produce 448 grams of Na2CO3 = (84.007×8.46) grams = 710.7 grams(answer)