Select the atom that will attract the electron pair (:) more strongly (the atom with the greater electronegativity)

Determine the limiting reactant (lr) and the mass (in g) of nitrogen that can be formed from 50.0 g n2o4 and 45.0 g n2h4. some p

Licemer1 [7]

N2O4(l) + 2 N2H4(l) → 3 N2(g) + 4 H2O(g)
1) to calculate the limiting reactant you need to pass grams to moles.
 moles is calculated by dividing mass by molar mass

mass of N2O4: 50.0 g
molar mass of N2O4 = 92.02 g/mol
molar mass of N2H4 = 32.05 g/mol.
mass of N2H4:45.0 g

moles N2O4=50.0/92.02 g/mol= 0,54 mol of N2O4
moles N2H4= 45/32.05 g/mol= 1,40 mol of N2H4

 2)
By looking at the balanced equation, you can see that 1 mol of N2O4 needs 2 moles of N2H4 to fully react . So to react 0,54 moles of N2O4, you need 2x0,54 moles of N2H4 moles
N2H4 needed = 1,08 moles.
You have more that 1,08 moles N2H4, so this means the limiting reagent is not N2H4, it's N2O4. The molecule that has molecules that are left is never the limiting reactant.

3) 1 mol of N2O4 reacting, will produce 3 mol of N2 (look at the equation)
There are 0,54 mol of N2O4 available to react, so how many moles will produce of N2?
1 mol N2O4------------3 mol of N2
0,54 mol N2O4--------x
x=1,62 mol of N2

4) the only thing left to do is convert the moles obtained, to grams.
We use the same formula as before, moles equal to mass divided by molar mass.
$moles= \frac{grams}{molar mass}$ (molar mass of N2= 28)
1,62 mol of N2= mass/ 28
mass of N2= 45,36 grams

4 0

3 years ago

A gardener waters the plants in a garden using a garden hose during a drought. The gardener observes that most of the water give

Anna71 [15]

Answer:

Break up the soil to increase the number of pores should be your answer.

Explanation:

If you do that it will increase the amount of water in the plant.

8 0

2 years ago

Read 2 more answers

The question is in the picture below

Rus_ich [418]

Answer:

$\Delta\text{H}_1+2\Delta\text{H}_2-\Delta\text{H}_3$

Explanation:

Hess's Law of Constant Heat Summation states that if a chemical equation can be written as the sum of several other chemical equations, the enthalpy change of the first chemical equation is equal to the sum of the enthalpy changes of the other chemical equations. Thus, the reaction that involves the conversion of reactant A to B, for example, has the same enthalpy change even if you convert A to C, before converting it to B. Regardless of how many steps it takes for the reactant to be converted to the product, the enthalpy change of the overall reaction is constant.

With Hess's Law in mind, let's see how A can be converted to 2C +E.

$\bf{\text{A} \rightarrow 2\text{B}}$ (Δ $\text{H}_1$ ) -----(1)

Since we have 2B, multiply the whole of II. by 2:

$\bf{2\text{B} \rightarrow 2\text{C} +2\text{D}}$ (2Δ $\text{H}_2$ ) -----(2)

This step converts all the B intermediates to 2C +2D. This means that the overall reaction at this stage is $\text{A} \rightarrow 2\text{C} +2\text{D}$ .

Reversing III. gives us a negative enthalpy change as such:

$\bf{2\text{D} \rightarrow \text{E}}$ (-Δ $\text{H}_3$ ) -----(3)

This step converts all the D intermediates formed from step (2) to E. This results in the overall equation of $\text{A} \rightarrow 2\text{C} +\text{E}$ , which is also the equation of interest.