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3 years ago

If 0.500 mol of acetylene is allowed to completely react with oxygen, what is the final yield of CO2 in moles?

Chemistry

1 answer:

Salsk061 [2.6K]3 years ago

4 0

Answer:

1.00 moles of CO2 is the final yield.

Explanation:

Based on the reaction:

2C2H2(g)+ 5O2(g) → 4CO2(g) + 2H2O(g)

<em>When 2 moles of acetylene (C2H2) completely reacts with oxygen, 4 moles of CO2 are produced.</em>

To solve this question we must use the chemical equation knowing: 2 moles C2H2 = 4 mol CO2

When 0.500 moles of acetylene react:

0.500 moles C2H2 * (4mol CO2 / 2mol C2H2) =

<h3>1.00 moles of CO2 is the final yield</h3>

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Burning 1 kg of coal releases about 3 million joules of energy. If you could use all of the chemical energy to lift another kilo

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I could lift 3.06 x 10⁵ m high

<h3>Further explanation </h3>

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3 years ago

For the mineral torbernite, Cu(UO2)2(PO4)2⋅8H2O, determine the ratio, by number, of H atoms to O atoms

astra-53 [7]

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4 years ago

Read 2 more answers

If 11.7 g of aluminum reacts with 37.2 g of copper (II) sulfate according to the following reaction, how many grams of aluminum

zhannawk [14.2K]

Answer:

There is 26.59 grams of aluminium sulfate produced

Explanation:

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

Mass of aluminium = 11.7 grams

Mass of copper (II) sulfate = 37.2 grams

Molar mass of Aluminium = 26.98 g/mol

Molar mass of CuSO4 = 159.61 g/mol

Molar mass of Al2(SO4)3 = 342.15 g/mol

<u />

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

2Al + 3CuSO4 → Al2(SO4)3 + 3Cu

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

Moles Al = mass Al / molar mass Al

Moles Al = 11.7 grams / 26.98 g/mol

Moles Al = 0.434 mol

<u>Step 4</u>: Calculate moles of CuSO4

Moles CuSO4 = 37.2 grams / 159.61 g/mol

Moles CuSO4 = 0.233 moles

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

For 2 moles of Al we need 3 moles of CuSO4

CuSO4 is the limiting reactant. It will be completely consumed (0.233 moles).

Al is in excess. There will be consumed 0.233 *(2/3) = 0.1553 moles

There will remain 0.434 - 0.1553 = 0.2787 moles

<u>Step 6: </u>Calculate moles of Al2(SO4)3

For 2 moles of Al we need 3 moles of CuSO4, to produce 1 mole of Al2(SO4)3 and 3 moles of Cu

For 0.233 moles CuSO4 we produce 0.233/3 = 0.0777 moles of Al2(SO4)3

<u>Step 7</u>: Calculate mass of Al2(SO4)3

Mass of Al2(SO4)3 = moles Al2(SO4)3 * molar mass Al2(SO4)3

Mass of Al2(SO4)3 = 0.0777 moles * 342.15g/mol

Mass of Al2(SO4)3 = 26.59 grams

There is 26.59 grams of aluminium sulfate produced

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3 years ago