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 possibly useful molar masses are as follows: N2O4 = 92.02 g/mol, N2H4 = 32.05 g/mol. N2O4(l) + 2 N2H4(l) → 3 N2(g) + 4 H2O(g)

A. The limiting reactant is N2O4

B. 45.67g

Explanation:

The balanced equation for the reaction is given below:

N2O4(l) + 2N2H4(l) → 3N2(g) + 4H2O(g)

A. To obtain the limiting reactant, first, let us calculate the mass of N2O4 and the mass of N2H4 that reacted from the balanced equation. This is illustrated below:

Molar Mass of N2O4 = 92.02 g/mol

Molar Mass of N2H4 = 32.05 g/mol

Mass of N2H4 that reacted from the balanced equation = 2 x 32.05 = 64.1g

Now we can determine the limiting reactant as follow:

Now, if we used all of the mass sample of N2O4 i.e 50g let us check to see if there will be left over for N2H4.

From the balanced equation above,

92.02g of N2O4 required 64.1g of N2H4.

Therefore, 50g of N2O4 will require = (50 x 64.1)/92.02 = 34.83g of N2H4.

Comparing the mass of N2H4 obtained ( i.e 34.83g) with the mass given ( i.e 45g) from the question, we can see that N2H4 is in excess as there would be a left over of (45 - 34.83 = 10.17g) og N2H4.

Therefore, N2O4 is the limiting reactant.

B. To obtain the mass of nitrogen formed, the limiting reactant will be used.

The equation for the reaction is given below:

N2O4(l) + 2N2H4(l) → 3N2(g) + 4H2O(g)

Molar Mass of N2 = 2 x 14.01 = 28.02g/mol

Mass of N2 from the balanced equation = 3 x 28.02 = 84.06g

Molar Mass of N2O4 = 92.02g/mol

From the balanced equation above,

92.02g of N2O4 produced 84.06g of N2.

Therefore, 50g of N2O4 will produce = (50 x 84.06) /92.02 = 45.67g of N2.

Therefore, 45.67g of nitrogen is formed from the reaction between 50.0g of N2O4 and 45.0g of N2H4

N2O4 is the limiting reactant

We will produce 45.6 grams of N2

Explanation:

Step 1: data given

Mass of N2O4 = 50.0 grams

Mass of N2H4 = 45.0 grams

Molar mass of N2O4 = 92.02 g/mol

Molar mass of N2H4 = 32.05 g/mol

Step 2: The balanced equation

N2O4(l) + 2 N2H4(l) → 3 N2(g) + 4 H2O(g)

Step 3: Calculate moles

Moles = mass / molar mass

Moles N2O4 = 50.0 grams / 92.02 g/mol

Moles N2O4 = 0.543 moles

Moles N2H4 = 45.0 grams / 32.05 g/mol

Moles N2H4 = 1.404 moles

Step 4: Calculate the limiting reactant

For 1 mol N2O4 we need 2 moles N2H4 to produce 3 moles N2 and 4 moles H2O

N2O4 is the limiting reactant. It will completely be consumed (0.543 moles). N2H4 is in excess. There will react 2*0.543 = 1.086 moles

There will remain 1.404 - 1.086 = 0.318 moles N2H4

Step 5: Calculate moles N2

For 1 mol N2O4 we need 2 moles N2H4 to produce 3 moles N2 and 4 moles H2O

For 0.543 moles N2O4 we'll have 3*0.543 = 1.629 moles N2

Step 6: Calculate mass N2

Mass N2 = moles N2 * molar mass N2

Mass N2 = 1.629 moles 28.0 g/mol

Mass N2 = 45.6 grams N2

We will produce 45.6 grams of N2

a is the answer to this question

explanation:

if you mean hydrogen and oxygen then water is the product