Answer:
sodium hydroxide is the limiting reactant
Explanation:
The first step is usually to put down the balanced reaction equation. This is the first thing to do when solving any problem related to stoichiometry. The balanced reaction equation serves as a guide during the solution.
2NBr3 + 3NaOH = N2 + 3NaBr + 3HOBr
Let us pick nitrogen gas as our product of interest. Any of the reactants that gives a lower number of moles of nitrogen gas is the limiting reactant.
For nitrogen tribromide
From the balanced reaction equation;
2 moles of nitrogen tribromide yields 1 mole of nitrogen gas
4.3 moles of nitrogen tribromide will yield 4.3 ×1/ 2 = 2.15 moles of nitrogen gas
For sodium hydroxide;
3 moles of sodium hydroxide yields 1 mole of nitrogen gas
5.9 moles of sodium hydroxide yields 5.9 × 1/ 3= 1.97 moles of nitrogen gas
Therefore, sodium hydroxide is the limiting reactant.
The balanced chemical reaction is:
Zn + 2AgNO3 = Zn(NO3)2 + 2Ag
To determine the amount of the reactant left, we have to determine which is the limiting and the excess reactant. We do as follows:
5.65 g Zn ( 1 mol / 65.38 g) = 0.09 mol Zn
21.6 g AgNO3 (1 mol / 169.87 g) = 0.13 mol AgNO3
The limiting reactant would be silver nitrate since it is consumed completely in the reaction. The excess reactant would be zinc.
Excess Zinc = 0.09 mol Zn - (0.13 / 2) mol Zn = 0.025 mol Zn left
Answer:
0.74 grams of methane
Explanation:
The balanced equation of the combustion reaction of methane with oxygen is:
it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.
firstly, we need to calculate the number of moles of both
for CH₄:
number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.
for O₂:
number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.
- it is clear that O₂ is the limiting reactant and methane will leftover.
using cross multiplication
1 mol of CH₄ needs → 2 mol of O₂
??? mol of CH₄ needs → 0.2812 mol of O₂
∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol
so 0.14 mol will react and the remaining CH₄
mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol
now we convert moles into grams
mass of CH₄ left over = no. of mol of CH₄ left over * molar mass
= 0.0469 mol * 16 g/mol = 0.7504 g
So, the right choice is 0.74 grams of methane
Independent and dependent variables and constants
Answer:
proactively develop ROE as well as responding to requests for additional ROE measures
Explanation:
The rules of engagement (roe) working group is a group who's main job is to proactively develop ROE as well as responding to requests for additional ROE measures. Rules of Engagement are a set or rules or directives for military soldiers that define the circumstances, conditions, degree, and manner in which the use of force can be used.
