Calcium nitrate will react with ammonium chloride at slightly elevated temperatures, as represented in the equation
Ca(NO3)2(s) + 2NH4Cl(s) ---> 2N2O(g) + CaCl2(s) + 4H2O(g)
What is the maximum volume of N2O at STP that could be produced using a 6.80-mol sample of each reactant?
Answer
0.007 L
1856 L
22.4 L
305 L
152 L
WEll, 1 mole of Calcium nitrate will yield 2 moles of N2O. Each mole of gas occupies 22.4 liters.
So 6.8*2*22.4=
22.4
To find the maximum volume of N2O at STP that could be produced, we need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed, thus limiting the amount of product that can be formed.
To find the limiting reactant, we can compare the moles of each reactant. The balanced equation tells us that 1 mole of Ca(NO3)2 reacts with 2 moles of NH4Cl to produce 2 moles of N2O.
For the given 6.80-mol sample of each reactant:
- The moles of Ca(NO3)2 is 6.80 mol.
- The moles of NH4Cl is 6.80 mol * 2 = 13.60 mol.
Since the stoichiometric ratio between Ca(NO3)2 and NH4Cl is 1:2, we have an excess of NH4Cl. Therefore, Ca(NO3)2 is the limiting reactant.
According to the balanced equation, 1 mole of Ca(NO3)2 produces 2 moles of N2O. Therefore, the maximum moles of N2O that can be produced is 6.80 mol * 2 = 13.60 mol.
Now we can use the ideal gas law to find the volume of N2O at STP (Standard Temperature and Pressure), which is 0 °C and 1 atm pressure. The molar volume of an ideal gas at STP is 22.4 L/mol.
The maximum volume of N2O at STP is:
Volume = moles of N2O * molar volume at STP
Volume = 13.60 mol * 22.4 L/mol
Volume = 305.6 L
Therefore, the maximum volume of N2O at STP that could be produced using a 6.80-mol sample of each reactant is 305 L. The correct answer is option D) 305 L.
To find the maximum volume of N2O at STP that could be produced, we need to use the concept of stoichiometry. Stoichiometry allows us to calculate the ratios of reactants and products in a chemical reaction.
First, we need to determine the limiting reactant in the reaction. The limiting reactant is the reactant that is completely consumed first, limiting the amount of product that can be formed.
We can use the balanced equation to determine the stoichiometric ratio between Calcium nitrate (Ca(NO3)2) and N2O. From the equation, we see that 1 mole of Ca(NO3)2 reacts to form 2 moles of N2O.
Therefore, the number of moles of N2O that can be produced from the 6.80 mol of Ca(NO3)2 is 6.80 mol x (2 mol N2O / 1 mol Ca(NO3)2) = 13.6 mol N2O.
Next, we can use the ideal gas law to calculate the volume of N2O at STP. The ideal gas law equation is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
At STP (standard temperature and pressure), the pressure is 1 atmosphere (atm) and the temperature is 273 Kelvin (K). We can rearrange the ideal gas law to solve for the volume:
V = (nRT) / P
Plugging in the values, we have:
V = (13.6 mol x 0.0821 L·atm/mol·K x 273 K) / 1 atm
V ≈ 305 L
Therefore, the maximum volume of N2O at STP that could be produced using a 6.80-mol sample of each reactant is approximately 305 L.
So, the correct answer is 305 L.