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Agata [3.3K]
3 years ago
5

For the following reaction, 22.6 grams of nitrogen monoxide are allowed to react with 4.64 grams of hydrogen gas . nitrogen mono

xide(g) + hydrogen(g) nitrogen(g) + water(l) What is the maximum amount of nitrogen gas that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams Submit AnswerRetry Entire Group
Chemistry
1 answer:
antiseptic1488 [7]3 years ago
3 0

Answer:

- 10.5 g of N₂

- Limiting reagent: NO

- 3.13 g of H₂ remains

Explanation:

First of all we state the reaction: 2NO(g) + 2H₂(g) → 2H₂O(l) + N₂(g)

We need to find out the limiting reactant and the excess reagent

Ratio in the reactants is 2:2. Let's convert the mass to moles:

22.6 g / 30 g/mol = 0.753 moles of NO

4.64 g / 2 g/mol = 2.32 moles of H₂

Certainly the limiting reagent is the NO and the excess reactant is the hydrogen:

- For 0.753 moles of NO, we need 0.753 moles of H₂ (we have 2.32 moles)

- For 2.32 moles of H₂, we need 2.32 moles of NO (and we don't have enough NO, because we only have 0.753 moles)

As the H₂ is the excess reagent, some moles still remains after the reaction is complete → 2.32 mol - 0.753 mol = 1.567 moles

We convert the moles to mass: 1.567 mol . 2g /1mol = 3.13 g of H₂ remains

As the NO is the limiting reagent, we can work with the equation:

We propose this rule of three: 2 moles of NO can produce 1 mol of N₂

Then, 0.753 moles of NO must produce (0.753 . 1) /2 = 0.376 moles of N₂

We convert the moles to mass 0.376 mol . 28 g / 1 mol = 10.5 g

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Answer:

-1.42, -0.375, 32.5% (.325), 3/8 (.375), √4 (2.0), 3 (3.0), 2³ (8.0)

8 0
1 year ago
Kepler's Law of Planetary Motion __________ the movement of planets in the universe
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3 years ago
Ammonia, NH3NH3 , can react with oxygen to form nitrogen gas and water. 4NH3(aq)+3O2(g)⟶2N2(g)+6H2O(l) 4NH3(aq)+3O2(g)⟶2N2(g)+6H
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Answer:

36.37% is the percent yield of the reaction.

Explanation:

4NH_3(aq)+3O_2(g)\rightarrow 2N_2(g)+6H_2O(l)

1)0.650 L nitrogen gas  , at 295 K and 1.01 bar.

Let the moles of nitrogen gas be n.

Pressure of the gas ,P=  1.01 bar = 0.9967 atm (1 bar = 0.9869 atm)

Temperature of the gas = T =  295 K

Volume of the gas = V = 0.650 L

Using an ideal gas equation:

PV=nRT

n=\frac{PV}{RT}=\frac{0.9967 atm\times 0.650 L}{0.0821 atm L/mol K\times 295 K}=0.0267 mol

2) Moles of ammonia gas=\frac{2.53 g}{17 g/mol}=0.1488 mol

Moles of oxygen gas =\frac{3.53 g}{32 g/mol}=0.1101 mol

According to reaction ,3 mol of oxygen reacts with 4 mol of ammonia.

Then,0.1101 mol of oxygen will react with:

\frac{4}{3}\times 0.1101 mol=0.1468 mol of ammonia.

Hence, oxygen gas is in limiting amount and act as limiting reagent.

3) Theoretical yield of nitrogen gas :

According to reaction, 3 mol of oxygen gas gives 2 moles of nitrogen gas.

Then 0.1101 mol of oxygen will give:

\frac{2}{3}\times 0.1101 mol=0.0734 mol of nitrogen.

Theoretical yield of nitrogen gas = 0.0734 mol

Experimental yield of nitrogen as calculated in part (1) = 0.0267 mol

Percentage yield:

\frac{\text{Experiential yield}}{\text{Theoretical yield}}\times 100

Percentage yield of the reaction:

\frac{ 0.0267 mol}{0.0734 mol}\times 100=36.37\%

36.37% is the percent yield of the reaction.

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2 years ago
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There are 2 gasses, A, B. They weigh 2.46g and 0.5g respectively, and the Volume of A is 3 times the volume of B. A has a molecu
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Answer:

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8 0
1 year ago
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