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Snowcat [4.5K]
2 years ago
14

If 5.85 moles of CaCO3 are used in an experiment, how many moles of carbon dioxide are created? (Be sure to include units in you

r answer.) CaCO3 → CaO + CO2
Chemistry
1 answer:
azamat2 years ago
7 0

Answer:

5.85 moles of carbon dioxide are created.

Explanation:

The balanced reaction is:

CaCO₃ → CaO + CO₂

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • CaCO₃: 1 mole
  • CaO: 1 mole
  • CO₂: 1 mole

Then you can apply the following rule of three: if by stoichiometry 1 mole of CaCO₃ produces 1 mole of CO₂, 5.85 moles of CaCO₃ will produce how many moles of CO₂?

moles of CO_{2}=\frac{5.85 moles of CaCO_{3} *1mole of CO_{2} }{1 mole of CaCO_{3}}

moles of CO₂= 5.85

<u><em>5.85 moles of carbon dioxide are created.</em></u>

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A 1.25 g gas sample occupies 663 ml at 25∘ c and 1.00 atm. what is the molar mass of the gas?
lakkis [162]

Using ideal gas equation,  

P\times V=n\times R\times T

Here,  

P denotes pressure  

V denotes volume  

n denotes number of moles of gas  

R denotes gas constant  

T denotes temperature  

The values at STP will be:  

P=1 atm  

T=25 C+273 K =298.15K

V=663 ml=0.663L

R=0.0821 atm L mol ⁻¹

Mass of gas given=1.25 g g

Molar mass of gas given=?

Number of moles of gas, n= \frac{Given mass of the gas}{Molar mass of the gas}

Number of moles of gas, n= \frac{1.25}{Molar mass of the gas}

Putting all the values in the above equation,

1\times 0.663=\frac{1.25}{Molar mass of the gas}\times 0.0821\times 298.15

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3 years ago
One of the intermediates in the synthesis of glycine from ammonia, carbon dioxide, and methane is aminoacetonitrile, C2H4N2. The
sleet_krkn [62]

Answer:

Mass of C₂H₄N₂ produced = 3.64 g

Explanation:

The balanced chemical equation for the reaction is given below:

3CH₄ (g) + 5CO₂ (g) + 8NH₃ (g) → 4C₂H₄N₂ (g) + 10H₂O (g)

From the equation, 3 moles of CH₄ reacts with 5 moles of CO₂ and 8 moles of NH₃ to produce 4 moles of C₂H₄N₂ and 10 moles of H₂O

Molar masses of the compounds are given below below:

CH₄ = 16 g/mol; CO₂ = 44 g/mol; NH3 = 17 g/mol; C₂H₄N₂ = 56 g/mol; H₂O g/mol

Comparing the mole ratios of the reacting masses;

CH₄ = 1.65/16 = 0.103

CO₂ = 13.5/44 = 0.307

NH₃ = 2.21/17 = 0.130

converting to whole number ratios by dividing with the smallest ratio

CH₄ = 0.103/0.103 = 1

CO₂ = 0.307/0.103 = 3

NH₃ = 0.130/0.103 = 1.3

Multiplying through with 5

CH₄ = 1 × 5 = 5

CO₂ = 3 × 5 = 15

NH₃ = 1.3 × 5 = 6.5

Therefore, the limiting reactant is NH₃

8 × 17 g (136 g) of NH₃ reacts to produce 4 × 56 g (224 g) of C₂H₄N₂

Therefore, 2.21 g of NH₃ will produce (2.21 × 224)/136 g of C₂H₄N₂ = 3.64 g of C₂H₄N₂

Mass of C₂H₄N₂ produced = 3.64 g

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