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k0ka [10]
3 years ago
9

Acetylene (C2H2) gas is often used in welding torches because of the very high heat produced when it reacts with oxygen (O2) gas

, producing carbon dioxide gas and water vapor. Calculate the moles of acetylene needed to produce 0.10 mol of carbon dioxide. Be sure your answer has a unit symbol
Chemistry
1 answer:
babymother [125]3 years ago
7 0

Answer:

0.05 mol

Explanation:

The balanced equation for the reaction that takes place is:

  • 2C₂H₂ (g) + 5O₂ (g) → 4CO₂ (g) + 2H₂O (g)

Now we<u> convert 0.10 moles of carbon dioxide (CO₂) into moles of acetylene (C₂H₂)</u>, using the <em>stoichiometric coefficients of the balanced reaction</em>:

  • 0.10 mol CO₂ * \frac{2molC_2H_2}{4molCO_2} = 0.05 mol C₂H₂
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Answer:

Explanation:

From the given information:

mass of silver chloride AgCl = 11.89 g

molar mass of AgCl = 143.37 g/mol

We know that:

number of moles = mass/molar mass

∴

number of moles of AgCl = 11.89 g/ 143.37 g/mol

number of moles of AgCl = 0.0829 mol

The chemical equation for the mineral called trona is:

\mathsf{Na_2CO_3.NaHCO_3.2H_2O}

when being reacted with hydrochloric acid, we have:

\mathsf{Na_2CO_3.NaHCO_3.2H_2O + 3HCl \to 3NaCl + 2CO_2 +4H_2O}

One mole of NaCl formed from one mole of trona sample = 0.0829 moles of AgCl

i.e. 0.0829 moles of NaCl can be formed from AgCl

mass of trona sample = number of moles × molar mass

mass of trona sample = 0.0829 × 226

mass of trona sample = 18.735 g

The mass in the percentage of NaHCO₃ = mass of NaHCO₃/ mass of trona

The mass in the percentage of NaHCO₃ = 6.93/18.735

The mass in the percentage of NaHCO₃ = 0.36989

The mass in the percentage of NaHCO₃ = 36.99%

Nonetheless, a 6.78 g samples manufactured from sodium carbonate in pure 100%

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6.78 g sample manufactured from Na₂CO₃ is purer.

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To answer this question, you will need to write the balanced equation and set up a BCA table. Using appropriate rounding rules,
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Water decomposes when electrolyzed to produce hydrogen and oxygen gas. If 2.5 grams of water were decomposed 1.04 grams of oxygen will be formed.

BCA table:

2H_{2}O ⇒ H_{2} + O_{2}

B  0.13        0 + 0

C  -0.13      0.065 + 0.065

A  0             0.065

Explanation:

Balanced equation for water decomposition into hydrogen and oxygen gases

   2H_{2}O ⇒ H_{2} + O_{2}

B  0.13        0 + 0

C  -0.13      0.065 + 0.065

A  0             0.065

Number of moles of water = \frac{mass}{atomic mass of 1 mole}

mass = 2.5 grams

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number of moles can be known by putting the values in the formula,

n = \frac{2.5}{18}

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2 moles of water gives one mole of oxygen on decomposition

so, 0.13 moles of water will give x moles of oxygen on decompsition

\frac{1}{2} = \frac{x}{0.13}

x = 0.065 moles of oxygen will be formed.

moles to gram will be calculated as

mass =number of moles x atomic mass

        = 0.065 x 16

         = 1.04 grams of oxygen.

7 0
4 years ago
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