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7. How many liters of NH3, at STP will react with 10.6 g O2 to form NO2 and water? 4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(g) *

Alexus [3.1K]

7) Answer is: c. 4.24 L.

Balanced chemical reaction: 4NH₃(g) + 7O₂(g) → 4NO₂(g) + 6H₂O(g).

m(O₂) = 10.6 g; mass of oxygen.

n(O₂) = m(O₂) ÷ M(O₂).

n(O₂) = 10.6 g ÷ 32 g/mol.

n(O₂) = 0.33 mol; amount of oxygen.

Vm = 22.4 L/mol; molar volume at STP (Standard Temperature and Pressure).

At STP one mole of gas occupies 22.4 liters of volume.

V(O₂) = n(O₂) · Vm.

V(O₂) = 0.33 mol · 22.4 L/mol.

V(O₂) = 7.42 L; volume of oxygen.

From balanced chemical reaction: n(O₂) : n(NH₃) = 7 : 4.

n(NH₃) = 4 · 0.33 mol ÷ 7.

n(NH₃) = 0.188 mol; amount of ammonia.

V(NH₃) = 0.188 mol · 22.4 L/mol.

V(NH₃) = 4.24 L.

8) Answer is: a. 3.7 g.

Balanced chemical reaction: P₄(g) + 6H₂(g) → 4PH₃(g).

m(P₄) = 3.4 g; mass of phosphorous.

n(P₄) = m(P₄) ÷ M(P₄).

n(P₄) = 3.4 g ÷ 123.9 g/mol.

n(P₄) = 0.0274 mol; limiting reactant.

n(H₂) = 4 g ÷ 2 g/mol.

n(H₂) = 2 mol; amount of hydrogen.

From balanced chemical reaction: n(P₄) : n(PH₃) = 1 : 4.

n(PH₃) = 4 · 0.0274 mol.

n(PH₃) = 0.1096 mol.

m(PH₃) = 0.1096 mol · 34 g/mol.

m(PH₃) = 3.726 g.

9) Answer is: d. Percent yield is the ratio of theoretical yield to actual yield expressed as a percent.

Percent yield = actual yield / theoretical yield.

Theoretical yield is the maximum amount of product that can be produced from limiting reactant and actual yield is a product that is obtained by experimentation.

For example:

the percent yield = 250 g ÷ 294.24 g · 100%.

the percent yield = 84.5 %.

3 0

3 years ago

What is the theoretical yield of Na2SO4, in grams?

Allushta [10]

Im sorry i dont know im sure someone will

3 0

3 years ago

Given these reactions, X ( s ) + 1 2 O 2 ( g ) ⟶ XO ( s ) Δ H = − 668.5 k J / m o l XCO 3 ( s ) ⟶ XO ( s ) + CO 2 ( g ) Δ H = +

qwelly [4]

Answer: The $\Delta H^o_{rxn}$ for the reaction is -1052.8 kJ.

Explanation:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

$X(s)+\frac{1}{2}O_2(g)+CO_2(g)\rightarrow XCO_3(s)$ $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $X(s)+\frac{1}{2}O_2(g)\rightarrow XO(s)$ $\Delta H_1=-668.5kJ$

(2) $XCO_3(s)\rightarrow XO(s)+CO_2$ $\Delta H_2=+384.3kJ$

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=[1\times \Delta H_1]+[1\times (-\Delta H_2)]$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-668.5))+(1\times (-384.3))=-1052.8kJ$

Hence, the $\Delta H^o_{rxn}$ for the reaction is -1052.8 kJ.

7 0

3 years ago

The surface tension of which of the following liquid is maximum?

Elden [556K]

The cohesive forces between liquid molecules are responsible for the phenomenon known as surface tension. I think the correct answer is option A. H2O will have the highest surface tension due to the hydrogen bonds that are present. Hope this answers the question. Have a nice day.

8 0

3 years ago

Which of the following is an accurate example of how the water cycle can influence local weather? A. An increase in evaporation

irina1246 [14]

The answer is; A

During a hot day, the land heats up faster than the waters. The air on land becomes warm and less dense fast and begin to rise in the atmosphere. The air on the ocean with is still cooler and denser moves in to replace the rising on land air. This causes a sea breeze. The sea breeze carries with it, moisture. The hotter the day the higher the humidity. When the air goes inland, it causes precipitation when it rises, cool, and condenses.

3 0

3 years ago

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