Answer:
O2 is the limiting reactant.
Explanation:
Step 1: Data given
Mass of NH3 = 2.00 grams
Mass of O2 = 2.50 grams
Molar mass NH3 = 17.03 g/mol
Molar mass O2 = 32 g/mol
Step 2: The balanced equation
4NH3(g) + 5O2 (g) → 4NO(g) + 6H2O (g)
Step 3: calculate moles NH3
Moles NH3 = mass NH3 / molar mass NH3
Moles NH3 = 2.00 grams / 17.03 g/mol
Moles NH3 = 0.117 moles
Step 4: Calculate moles O2
Moles O2 = mass / molar mass O2
Moles O2 = 2.50 grams / 32 g/mol
Moles O2 = 0.0781 moles
Step 5: Calculate the limiting reactant
For 4 moles NH3 we need 5 moles O2 to produce 4 moles NO and 6 moles H2O
O2 is the limiting reactant. It will completely be consumed. (0.0781 moles). NH3 is in excess. There will react 4/5 * 0.0781 moles = 0.0625 moles
There will remain 0.117 - 0.0625 = 0.0545 moles NH3
O2 is the limiting reactant.
Answer:
Visible light waves
Explanation:
Visible light waves are the only electromagnetic waves we can see. We see these waves as the colors of the rainbow. Each color has a different wavelength. Red has the longest wavelength and violet has the shortest wavelength.
Answer:
Cancel out CO because it appears as a reactant in one intermediate reaction and a product in the other intermediate reaction.
Explanation:
The CO appears twice hence in he intermediate reaction it only forms path of the enabling reagents and it further reacts to form the final product. Accounting for the CO in the intermediate reaction that undergoes further reaction will impact on the stoichiometry of the reaction.
Answer:
Dissolve 226 g of KCl in enough water to make 1.5 L of solution
Explanation:
1. Calculate the moles of KCl needed
2. Calculate the mass of KCl
3. Prepare the solution
- Measure out 224 g of KCl.
- Dissolve the KCl in a few hundred millilitres of distilled water.
- Add enough water to make 1.5 L of solution.
Mix thoroughly to get a uniform solution.
You're looking for the number of moles of H2, and you have 6.0 mol Al and 13 mol HCL.
For the first part, you have to make your way from 6.0 mol of Al to mol of H2, right? For that to happen, you need to make a conversion factor that will cancel the mol Al, in such case use the 2 moles of Al from your equation to cancel them out. At the top of the equation, you can use the number of moles of H2 from the equation and find the moles that will be produced for the H2.
6.0mol Al x 3 mol H2/2 mol Al = 9 mol H2
For the second part, you have to make the same procedure, make a conversion factor that will cancel the mol of HCL and for that you need to use the 6 mol HCL from your equation, and at the numerator you can put the 3 mol of H2 from the equation so that you can find the number of moles of H2 that will be produced.
13 mol HCL x 3 mol H2/6 mol HCL = 6.5 mol H2
As it can be seen, HCL produces the less amount of H2 moles. Therefore, the reaction CANNOT produce more than 6.5 mol H2, in that case 6.5 mol will be the maximum number of moles that will be produced at the end because HCL does not have enough to produce more than 6.5 mol.
In that case HCL is the limiting reactant because it limits that will be produced, and so the answer is B!
