Question

8.03 x 10^23 moleucles of oxygen react with excess nitrogen. what is the maximum number of grams of dinitrogen tetroxide that can be produced?

42 grams of cyclohexane burns in excess air to form carbon dioxide and water. How many grams of carbon dioxide and of water vapor are produced?

Answer 1

We know that 1 mole of any substance contains 6.023x10 ²³ ( avagadro number ) molecules

                   1 mole of O₂ contains 6.023x10 ²³ molecules

                   Z moles of O₂ contains 8.03x10 ²³ molecules

                         

The balanced equation is         2O₂ + N₂ N₂O₄

According to the above balanced equation ,

2 moles of O₂ produces 1 mole of N₂O₄

1.33 moles of O₂ produces M mole of N₂O₄

                      

Molar mass of N₂O₄ is = (2x14) + (4x16) = 92 g/mol

So mass of N₂O₄ produced is , m = Number of moles x molarmass

                                                   = 0.665 mol x 92 g/mol

                                                   = 61.2 g

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The balanced equation is      C₆H₁₂ + 9O₂ 6CO₂ + 6H₂O

Molar mass of C₆H₁₂ is = (6x12) + (12x1) = 84 g/mol

Molar mass of CO₂ is = 12 + (2x16) = 44 g/mol

Molar mass of H₂O is = (2x1) + 16 = 18 g/mol

According to the balanced equation ,

84 g of C₆H₁₂ produces 6x44 g of CO₂

84/2 = 42 g of C₆H₁₂ produces (6x44) / 2 = 132 g of CO₂

From the balanced equation ,

84 g of C₆H₁₂ produces 6x18 g of H₂O

84/2 = 42 g of C₆H₁₂ produces (6x18) / 2 = 54 g of H₂O