Question

Assume that coal can be represented by the chemical formula C135H96O9NS. If 4.0 tons of coal is burned, what mass of nitrogen is found in the NO produced by this process?

Answer 1

Answer:

26653.9 grams of nitrogen is found in the nitrogen dioxide produced by this process.

Explanation:

$C_{135}H_{96}O_9NS + \frac{313}{2}O_2\rightarrow 135CO_2 + 48H_2O+ NO_2 + SO_2$

Mass of the coal = 4.0 tons  

1 ton = 907185 g

$4.0 tons = 4\times 907185g=3,628,740 g$

Molar mass of coal = 1,906 g/mol

Moles of coals = $\frac{3,628,740 g}{1,906 g/mol}=1,903.85 mol$

According to reaction,  1 mole of coal gives 1 mole of nitrogen dioxide gas.

Then, 1,903.85 mole of coal on combustion will give:

$\frac{1}{1}\times 1,903.85 mol=1,903.85 mol$ of nitrogen dioxide gas

1 mole of nitrogen dioxide gas has 1 mole of nitrogen atom.Then 1,903.85 moles of nitrogen dioxide will have ;

$1\times 1,903.85 mol=1,903.85 mol$ of nitrogen

Mass of 1,903.85 moles of nitrogen dioxide gas :

1,903.85 mol × 14 g/mol = 26653.9 g

26653.9 grams of nitrogen is found in the nitrogen dioxide produced by this process.

Answer 2

26656 g of nitrogen is produced when 4.0 tons of coal is burned.

The reaction equation ought to be;

$C135H96O9NS + 156O2 --------> 135CO2 + 48H2O + NO + SO2$

Mass of 4.0 tons of coal in gram = 3.629 × 10^6 g

Molar mass of C135H96O9NS = 135(12) + 96(1) + 9(16) + 14 + 32

= 1620 + 96 + 144 + 14 + 32 = 1906 g/mol

Number of moles of coal burned = 3.629 × 10^6 g/ 1906 g/mol = 1904 moles of coal

If 1 mole of coal produced 1 mole of NO

1904 moles of coal produces 1904 moles of NO

Mass of NO produced = 1904 moles × 30 g/mol = 57120 g

1 mole of NO = 30 g of NO

If 14 g of nitrogen is contained in 30 g of NO

xg of nitrogen is contained in 57120 g of NO

x = 14 g ×57120 g of NO/30 g

x = 26656 g of nitrogen

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