Answer:
For Mass N, Mass H, and Mass O, the mass is 28.0 g N, 4.0 g H, and 48.0 g respectively
Explanation:
The computation of the mass of each element is given below:
As we know that
A1 mole of ammonium nitrate i.e. 2 mol N, 4 mol H, 3 mol
Now we multiply each of above by the molar masses
For N
= 14.0 g/mol × 2
= 28.0 gN
For H
= 1.0 g/mol × 4
= 4.0 gN
ANd, for O
= 16.0 g/mol × 3
= 48.0 gN
Hence, For Mass N, Mass H, and Mass O, the mass is 28.0 g N, 4.0 g H, and 48.0 g respectively
Answer:
Limiting reactant = B2O3
Amount of BCl3 formed = 468 g
Explanation:
The given reaction is:
In order to identify the limiting reagent calculate the moles of B2O3, C and Cl2. The reagent with the lowest moles is the limiting reactant
Since the moles of B2O3 < C < Cl2, the limiting reactant is B2O3
Based on the reaction stoichiometry:
1 mole of B2O3 produces 2 moles of BCl3
Hence, the number of moles of BCl3 produced under the experimental conditions = 2*1.997=3.994 moles