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Temka [501]
2 years ago
10

In an experiment, 3.25 g of NH3 are allowed to react with 3.50 g of O2. 4NH3 + 5O2 > 4NO + 6H2O a. Which reactant is the limi

ting reagent? O2 b. How many grams of NO are formed
Chemistry
1 answer:
cluponka [151]2 years ago
7 0

Answer:

2.61 g of NO will be formed

The limiting reagent is the O₂

Explanation:

The reaction is:

4NH₃  +  5O₂  →  4NO  +  6H₂O

We convert the mass of the reactants to moles:

3.25g / 17 g/mol = 0.191 moles of NH₃

3.50g / 32 g/mol =0.109 moles of O₂

Let's determine the limiting reactant by stoichiometry:

4 moles of ammonia react with 5 moles of oxygen

Then, 0.191 moles of ammonia will react with (0.191 . 5) / 4 = 0.238 moles of oxygen. We only have 0.109 moles of O₂ and we need 0.238, so as the oxygen is not enough, this is the limiting reagent

Ratio with NO is 5:4

5 moles of oxygen produce 4 moles of NO

0.109 moles will produce (0.109 . 4)/ 5 = 0.0872 moles of NO

We convert the moles to mass, to get the answer

0.0872 mol . 30g / 1 mol = 2.61 g

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Answer:

it is 1,4

Explanation:

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2 years ago
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Describe and explain how petrol is separated from the mixture of hydrocarbons in crude oil?
leonid [27]
It's an exciting and interesting process but those 5 points aren't worth a decent answer. The whole process is called fractioned distillation where different temperatures are reached, transforming the crude oil in gas. After being hit by a cold source it colds becoming different hydrocarbons depending on the temperatures.
6 0
3 years ago
How much heat is required to vaporize 31.5 gg of acetone (C3H6O)(C3H6O) at 25 ∘C∘C? The heat of vaporization for acetone at this
KiRa [710]

Answer:

≅ 16.81 kJ

Explanation:

Given that;

mass of acetone = 31.5 g

molar mass of acetone = 58.08 g/mol

heat of vaporization for acetone = 31.0 kJ/molkJ/mol.

Number of moles = \frac{mass}{molar mass}

Number of moles of acetone = \frac{31.5}{58.08}

Number of moles  of acetone = 0.5424 mole

The heat required to vaporize 31.5 g of acetone can be determined by multiplying the number of moles of acetone with the heat of vaporization of acetone;

Hence;

The heat required to vaporize 31.5 g of acetone = 0.5424 mole × 31.0 kJ/mol

The heat required to vaporize 31.5 g of acetone = 16.8144 kJ

≅ 16.81 kJ

4 0
3 years ago
If 4.65 LL of CO2CO2 gas at 22 ∘C∘C at 793 mmHg mmHg is used, what is the final volume, in liters, of the gas at 35 ∘C∘C and a p
otez555 [7]

Answer:

About 7.9 L.

Explanation:

We can utilize the ideal gas law. Recall that:

\displaystyle PV = nRT

Because the amount of carbon dioxide does not change, we can rearrange to formula to:
\displaystyle \frac{PV}{T}= nR

Because the right-hand side stays constant, we have that:
\displaystyle \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} = nR

Hence substitute initial values and known final values:
\displaystyle \begin{aligned} \frac{(793\text{ mm Hg})(4.65 \text{ L})}{(22 \text{ $^\circ$C})} & = \frac{(743 \text{ mm Hg})V_2}{(35\text{ $^\circ$C})} \\ \\ V_2 & = 7.9\text{ L}\end{aligned}

Therefore, the final volume is about 7.9 L.

8 0
2 years ago
Silver nitrate reacts with calcium to make calcium nitrate and silver. The reaction is exothermic and produces 18.7 kJ/mol of en
denis-greek [22]

Answer:

1.63 kilo Joules of energy will be produced when 14.8 g of silver nitrate react.

Explanation:

AgNO_3+Ca\rightarrow Ag+CaNO_3,\Delta H_{rxn}=-18.7 kJ/mol

Mass of silver nitrate = 14.8 g

Moles of silver nitrate = \frac{14.8 g}{170 g/mol}=0.08706 mol

According to reaction, when 1 mole of silver nitrate reacts with 1 mole of calcium it gives 18.7 kJ of heat.

Then amount of heat released when 0.08706 moles of silver nitrate reacts :

0.08706\times 18.7 kJ=1.628 kJ\approx 1.63 kJ

1.63 kilo Joules of energy will be produced when 14.8 g of silver nitrate react.

6 0
2 years ago
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