In: Chemistry
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?
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