1. A sample of 10.7 g of CO reacts completely with O2 to form CO2. How many grams of CO2 will be formed?
2. Urea (NH2)2CO is prepared by reacting ammonia with carbon dioxide. The byproduct is water. 637.2 g of ammonia are reacted with 787.3 g of carbon dioxide. Which of the two reactants is a limiting reactant.
Convert each reactant to moles, then use the coefficients in the balanced equation to convert EACH reactant to moles of the product. The two answers can't be right; the correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
To answer both of these questions, we need to use the concept of stoichiometry in chemical reactions. Stoichiometry allows us to determine the relative quantities of reactants and products involved in a chemical reaction.
1. To determine the mass of CO2 formed, we should first write a balanced chemical equation for the reaction between CO and O2:
2CO + O2 -> 2CO2
From the balanced equation, we can see that 2 moles of CO react with 1 mole of O2 to produce 2 moles of CO2. The molar mass of CO2 is approximately 44.01 g/mol.
Now, let's calculate the molar mass of CO to determine the number of moles present in 10.7 g of CO:
Molar mass of CO = 12.01 g/mol (carbon) + 16.00 g/mol (oxygen) = 28.01 g/mol
Number of moles of CO = mass / molar mass = 10.7 g / 28.01 g/mol ≈ 0.382 mol
Since the balanced equation tells us that 2 moles of CO react to form 2 moles of CO2, we can conclude that 0.382 moles of CO will produce 0.382 moles of CO2.
Finally, let's calculate the mass of CO2 formed:
Mass of CO2 = moles of CO2 × molar mass of CO2 = 0.382 mol × 44.01 g/mol ≈ 16.83 g
Therefore, approximately 16.83 grams of CO2 will be formed.
2. To determine the limiting reactant in the reaction between ammonia (NH3) and carbon dioxide (CO2), we once again need to write a balanced chemical equation:
2NH3 + CO2 -> (NH2)2CO + H2O
From the balanced equation, we can see that 2 moles of NH3 react with 1 mole of CO2 to produce 1 mole of (NH2)2CO.
To determine the limiting reactant, we need to compare the number of moles of each reactant with the stoichiometric ratio given by the balanced equation.
Number of moles of NH3 = mass of NH3 / molar mass of NH3
= 637.2 g / 17.03 g/mol ≈ 37.5 mol
Number of moles of CO2 = mass of CO2 / molar mass of CO2
= 787.3 g / 44.01 g/mol ≈ 17.88 mol
Now, let's compare the moles of NH3 to the moles of CO2 for their stoichiometric ratio.
Moles ratio of NH3 to CO2 = 2 mol NH3 / 1 mol CO2
= 2 mol NH3 / (2 × 1 mol NH3) = 1
Since the mole ratio of NH3 to CO2 is 1, it means that one mole of NH3 is required for one mole of CO2. Therefore, the limiting reactant is NH3.
In conclusion, ammonia (NH3) is the limiting reactant in the given reaction.