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2 years ago

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How many moles of nitrogen gas would be produced if 4.92 moles of copper(II) oxide were reacted with excess ammonia in the follo

wing chemical reaction? 2 NH₃(g) + 3 CuO (s) → 3 Cu(s) + N₂(g) + 3 H₂O(g)\

1 answer:

WITCHER [35]2 years ago

3 0

Answer:

4.92/3

Explanation:

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The reaction, 2C4H10 (g) + 13O2 (g) à 8CO2 (g) + 5H2O (g), is the combustion of butane. What occurs as the reaction proceeds?

Solnce55 [7]

The reaction, 2 C4H10 (g) + 13 O2 (g) = 8 CO2 (g) + 5 H2O (g), is the combustion of butane. A combustion reaction involves the reaction of a hydrocarbon with oxygen producing carbon dioxide and water. This reaction is exothermic which means it releases energy in the form of heat. Therefore, as the reaction proceeds,a heat energy is being given off by the reaction. This happens because the total kinetic energy of the reactants is greater than the total kinetic energy of the products. So, the excess energy should be given off somewhere which in this case is released as heat.

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3 years ago

Who started a colony for the Quakers? A) John Smith B) Lord Baltimore C) William Penn D) Sir George Carteret

vovikov84 [41]

Answer:

William Penn

Explanation:

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3 years ago

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skelet666 [1.2K]

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4 years ago

In which reaction does the oxidation number of hydrogen change? In which reaction does the oxidation number of hydrogen change?

dedylja [7]

<u>Answer:</u> The correct answer is $2Na(s)+2H_2O(l)\rightarrow 2NaOH(aq.)+H_2(g)$

<u>Explanation:</u>

Oxidation number is defined as the number which is given to an atom when it looses or gains electron. When an atom looses electron, it attains a positive oxidation state. When an atom gains electron, it attains a negative oxidation state.

Oxidation state of the atoms in their elemental state is considered as 0. Hydrogen is present as gaseous state.

For the given chemical reactions:

<u>Reaction 1:</u> $2HClO_4(aq.)+CaCO_3(s)\rightarrow Ca(ClO_4)_2(aq.)+H_2O(l)+CO_2 (g)$

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: +1

Thus, the oxidation state of hydrogen is not changing.

<u>Reaction 2:</u> $CaO(s)+H_2O(l)\rightarrow Ca(OH)_2(s)$

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: +1

Thus, the oxidation state of hydrogen is not changing.

<u>Reaction 3:</u> $HCl(aq.)+NaOH(aq.)\rightarrow NaCl(aq.)+H_2O(l)$

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: +1

Thus, the oxidation state of hydrogen is not changing.

<u>Reaction 4:</u> $2Na(s)+2H_2O(l)\rightarrow 2NaOH(aq.)+H_2(g)$

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: 0

Thus, the oxidation state of hydrogen is changing.

<u>Reaction 5:</u> $SO_2(g)+H_2O(l)\rightarrow H_2SO_3(aq.)$

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: +1

Thus, the oxidation state of hydrogen is not changing.

Hence, the correct answer is $2Na(s)+2H_2O(l)\rightarrow 2NaOH(aq.)+H_2(g)$

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3 years ago

Urea (CH4N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH3) with carbon dioxide as follows: 2N

Gnesinka [82]

Answer:

NH3 is the limiting reactant

The theoretical yield is 216.0 kg urea

The % for this reaction is 78.8 %

Explanation:

<u>Step 1:</u> Data given

Mass of ammonia = 122.5 kg

Mass of carbon dioxide = 211.4 kg

Mass of urea produced = 170.3 kg

Molar mass of ammnoia = 17.031 g/mol

Molar mass of carbon dioxide = 44.01 g/mol

Moalr mass of urea = 60.06 g/mol

<u>Step 2:</u> The balanced equation

2NH3(aq) + CO2(aq) --> CH4N2O(aq) + H2O(l)

<u>Step 3:</u> Calculate moles of NH3

Number of moles = mass / Molar mass

Moles NH3 = 122500 grams / 17.031 g/mol

Moles NH3 = 7192.77 moles

<u>Step 4:</u> Calculate moles of CO2

Moles CO2 = 211400 / 44.01 g/mol

Moles CO2 = 4803.45 moles

<u>Step 5</u>: Calculate limiting reactant

For 2 moles NH3 consumed, we need 1 moles of CO2 to produce 1 mole urea and 1 mole H2O

NH3 is the limiting reactant. It will completely be consumed (7192.77 moles).

CO2 is in excess. There will be consumed 7192.77/2 = 3596.4 moles

There will remain 4803.45 - 3596.4 = 1207.05 moles of CO2

<u>Step 6:</u> Calculate moles of urea produced:

For 2 moles NH3 consumed, we need 1 moles of CO2 to produce 1 mole urea and 1 mole H2O

For 7192.77 moles of NH3, we have 3596.4 moles of urea produced

<u>Step 7: </u>Calculate mass of urea

Mass urea = moles urea * molar mass urea

Mass urea = 3596.4 moles * 60.06 g/mol

Mass urea = 216000 grams = 216 kg = theoretical yield

<u>Step 8</u>: Calculate % yield

% yield = (actual yield / theoretical yield)*100%

% yield = (170.3 / 216) *100% = 78.8%

The % for this reaction is 78.8 %

3 0

3 years ago