4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(g)For the balanced reaction above, how many grams of nitrogen monoxide are produced when 16.0 g of oxygen react?

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11ecba2d_124c_07c0_a3bf_69d4972a2996_MD0008_00 The molar mass of a compound is the amount of mass contained in 1 mole of the compound (grams per mole) and can be used to convert the mass of a compound to the number of moles. The reactants and/or products of a particular reaction are related to one another by the mole ratios (stoichiometry) from a balanced reaction equation or chemical formula. For the reaction in this question, if 16.0 g of O₂ react, the number of moles of O₂ participating in the reaction can be calculated by dividing the mass of O₂ by its molar mass to give 0.500 mol O₂. The balanced reaction equation shows that 4 moles of NO are produced for every 5 moles of O₂ that react. When this stoichiometric ratio is applied, the number of moles of NO produced from 0.500 mol O₂ can be found: 0.400 mol NO. The mass of the 0.400 mol of NO produced in the reaction can be determined by multiplying the moles by the molar mass of NO (30.01 g/mol). Therefore, 12.0 g of nitrogen monoxide are produced when 16.0 g of oxygen react. (Choice B) 13.6 g is obtained if the numerical values in molar mass ratios are inverted (as in 32.00 mol/g and 30.01 mol/g) during the first and last steps of the calculation. (Choice C) 18.8 g is obtained if the stoichiometric mole ratio is inverted during the second step of the calculation. (Choice D) 24.0 g is obtained if 16.00 g/mol (the molar mass of an oxygen atom), is used instead of 32.00 g/mol (the molar mass of an oxygen molecule) during the first step of the calculation. Educational objective: The molar mass (the amount of mass in 1 mole of a compound) permits conversions between a given amount of mass and the number of moles that it contains. Stoichiometric mole ratios from a balanced reaction equation can be applied to determine the number of moles of one chemical species relative to the moles of another chemical species in a reaction.

4 NH3(g) + 5 O2(g) → 4 NO(g) + 6

4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6