CHEMISTRY HIGH SCHOOL

60/1 point
If you have 1.0 mole of each of the following compounds, which will have the most molecules?
Х
NI3 (nitrogen triiodide)
KCI (potassium chloride)
Lici (lithium chloride)
O C2H2 (ethyne)
O They all have the same number of molecules.
7
1/1 point
fn 35 moles of Iron (Fe)?

Answers

Answer 1

Answer:

6. They all have the same number of molecules.

7. 19.5 g

Explanation:

6. From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02×10²³ molecules i.e

1 mole = 6.02×10²³ molecules

Considering the question given above, each compounds has just 1 mole. Thus, they will also have the same number of molecules.

7. Determination of the mass of iron (Fe)

Number of mole of Fe = 0.35 mole

Molar mass of Fe = 55.85 g/mol

Mass of Fe =?

Mass = mole × molar mass

Mass of Fe = 0.35 × 55.85

Mass of Fe = 19.5 g

Thus, the mass of 0.35 mole of iron (Fe) is 19.5 g

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A gas contains a mixture of NH3(g) and N2H4(g), both of which react with O2(g) to form NO2(g) and H2O(g). The gaseous mixture (with an initial mass of 61.00 g) is reacted with 10.00 moles O2, and after the reaction is complete, 4.062 moles O2 remains. Calculate themass percent of N2H4(g) in the original gaseous mixture.

Answers

Answer:

Mass percent of N2H4 in original gaseous mixture = 31.13 %

Explanation:

Given:

Initial mass of gaseous mixture = 61.00 g

Initial mole of oxygen = 10.0 mol

Moles of oxygen remaining after the reaction = 4.062 mol

Moles of oxygen used = 10.0 - 4.062 = 5.938 mol

$4NH_3 + 7O_2\rightarrow 4NO_2 + 6H_2O$

$N_2H_4 + 3O_2\rightarrow 2NO_2 + 2H_2O$

Total oxygen used in both the reactions = 10.0 parts

out of 10 parts, 3 part react with N2H4.

$No.\;of\;moles\;of \;oxygen \;used = 5.398*(3)/(10) =1.78\; moles$

Now, consider the reaction of N2H4

$N_2H_4 + 3O_2\rightarrow 2NO_2 + 2H_2O$

3 moles of O2 react with 1 mole of N2H4

1.78 moles of oxygen will react with 1.78/3 = 0.5933 mol of N2H4

$Mass = Moles* Molecular mass$

Molecular mass of N2H4 = 32 g/mol

$Mass\;of\;N_2H_4= 0.5933* 32 = 18.99 g$

$Mass\;percent = (Mass\;of\;N_2H_4)/(Total\;mass)* 100$

Total mass = 61.0 g

$Mass\;percent = (18.99)/(61.0)* 100=31.13 \%$

Final answer:

The mass percent of N2H4 in the gaseous mixture can be determined through stoichiometric calculations and determining the limiting reactant. The initial and remaining amounts of O2 are used to calculate the reacted amount of O2, which then allows for the calculation of the amount of N2H4. This information is used in the mass percent formula.

Explanation:

The balanced reaction states that for one mole of NH3, one mole of O2 is required, while for one mole of N2H4, 3 moles of O2 are required. Thus, the initial moles of O2 were 10 moles and after reaction 4.062 moles O2 remained. Thus, the reacted amount of O2 is 10 - 4.062 = 5.938 moles. From calculating the limiting reactant and applying stoichiometry, the amount of N2H4 can be determined. We know the molar mass of N2H4 is 32 g/mole. By calculating the molar ratio, we can then calculate the mass percent of N2H4 in the mixture using the formula: (mass of N2H4 / total mass) * 100%.