Consider the following reaction at 1.10 atm and 19°C:
2NaCL+2NH3+CO2+H2O->2NH4CL+Na2CO3
0.216 mol of sodium chloride, 2.98 L of ammonia, 2.00 L of carbon dioxide, and an unlimited amount of water react to form aqueous ammonium chloride and solid sodium bicarbonate. How many moles of ammonium chloride are formed in the reaction?
a. 0.137 mol
b. 0.274 mol
c. 0.216 mol
d. 5.96 mol
e. 0.0918 mol
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To determine the moles of ammonium chloride formed in the reaction, we need to first determine the limiting reagent. The limiting reagent is the reactant that will be completely consumed in the reaction, and it restricts the amount of product that can be formed.
First, we need to calculate the moles of each reactant:
moles of sodium chloride = 0.216 mol
moles of ammonia = (2.98 L)(1 mol/22.4 L) = 0.133 mol
moles of carbon dioxide = (2.00 L)(1 mol/22.4 L) = 0.089 mol
According to the balanced equation, the stoichiometric ratio between sodium chloride and ammonium chloride is 2:2. Therefore, the moles of ammonium chloride formed will be the same as the moles of sodium chloride, which is 0.216 mol.
Therefore, the correct answer is:
c. 0.216 mol
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