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jeyben [28]
3 years ago
7

Hydrazine (n2h4) is used as rocket fuels. it reacts with oxygen to form nitrogen and water. write the balanced equation for this

reaction. n2h4 + o2 → n2 +2h2o how many liters of n2 at stp form when 100g of n2h4 reaction with 100g of o2? how many grams of the excess reagent remain after the reaction?
Chemistry
1 answer:
FrozenT [24]3 years ago
5 0

The answer is: volume of nitrogen is 70L.

Chemical reaction: N₂H₄ + O₂ → N₂ + 2H₂O.

m(N₂H₄) = 100 g; mass of hydrazine.

M(N₂H₄) = 32 g/mol; molar mass of hydrazine.

n(N₂H₄) = m(N₂H₄) ÷ M(N₂H₄).

n(N₂H₄) = 100 g ÷ 32 g/mol.

n(N₂H₄) = 3.125 mol; amount of hydrazine.

m(O₂) = 100 g; mass of oxygen.

M(O₂) = 32 g/mol; molar mass of oxygen.

n(O₂) = 100 g ÷ 32 g/mol.

n(O₂) = 3.125 mol; amount of oxygen.

From chemical reaction: n(O₂) : n(N₂) = 1 : 1; n(O₂) = n(N₂).

n(N₂) = 3.125 mol; amount og nitrogen gas.

V(N₂) = n(N₂) · Vm.

Vm = 22.4 L/mol; molar volume.

V(N₂) = 3.125 mol · 22.4 L/mol.

V(N₂) = 70 L.

There is not excess reagent, because hydrazine and oxygen are all used in chemical reaction.

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if i add 25 ml of water to 135 ml of a 0.25 M NaOH solution what will the molarity of the diluted solution be​
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Answer:

0.21 M. (2 sig. fig.)

Explanation:

The molarity of a solution is the number of moles of the solute in each liter of the solution. The unit for molarity is M. One M equals to one mole per liter.

How many moles of NaOH in the original solution?

n = c \cdot V,

where

  • n is the number of moles of the solute in the solution.
  • c is the concentration of the solution. c = 0.25 \;\text{M} = 0.25\;\text{mol}\cdot\textbf{L}^{-1} for the initial solution.
  • V is the volume of the solution. For the initial solution, V = 135\;\textbf{mL} = 0.135\;\textbf{L} for the initial solution.

n = c\cdot V = 0.25\;\text{mol}\cdot\textbf{L}^{-1} \times 0.135\;\textbf{L} = 0.03375\;\text{mol}.

What's the concentration of the diluted solution?

\displaystyle c = \frac{n}{V}.

  • n is the number of solute in the solution. Diluting the solution does not influence the value of n. n = 0.03375\;\text{mol} for the diluted solution.
  • Volume of the diluted solution: 25\;\text{mL} + 135\;\text{mL}  = 160\;\textbf{mL} = 0.160\;\textbf{L}.

Concentration of the diluted solution:

\displaystyle c = \frac{n}{V} = \frac{0.03375\;\text{mol}}{0.160\;\textbf{L}} = 0.021\;\text{mol}\cdot\textbf{L}^{-1} = 0.021\;\text{M}.

The least significant number in the question comes with 2 sig. fig. Keep more sig. fig. than that in calculations but round the final result to 2 sig. fig. Hence the result: 0.021 M.

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