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3 years ago

2C₂H6 + 702 —>4C02 + 6H₂O

Chemistry

1 answer:

navik [9.2K]3 years ago

8 0

Answer:

975.56×10²³ molecules

Explanation:

Given data:

Number of molecules of C₂H₆ = 4.88×10²⁵

Number of molecules of CO₂ produced = ?

Solution:

Chemical equation:

2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

Number of moles of C₂H₆:

1 mole = 6.022×10²³ molecules

4.88×10²⁵ molecules×1mol/6.022×10²³ molecules

0.81×10² mol

81 mol

Now we will compare the moles of C₂H₆ with CO₂.

C₂H₆ : CO₂

2 : 4

81 : 4/2×81 = 162 mol

Number of molecules of CO₂:

1 mole = 6.022×10²³ molecules

162 mol ×6.022×10²³ molecules / 1 mol

975.56×10²³ molecules

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<u>Answer:</u> 0.0237 g of calcium carbonate would be required to neutralize the given amount of HCl

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$pH=-\log [H^+]$ .....(1)

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$\text{Molarity of solution}=\frac{\text{Number of moles of solute}\times 1000}{\text{Volume of solution (mL)}}$ .....(2)

We are given:

Volume of solution = 15.0 mL

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Putting values in equation 2:

$0.0316=\frac{\text{Moles of HCl}\times 1000}{15.0}\\\\\text{Moles of HCl}=\frac{0.0316\times 15.0}{1000}=4.74\times 10^{-4}mol$

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$2HCl+CaCO_3\rightarrow H_2CO_3+CaCl_2$

By the stoichiometry of the reaction:

2 moles of HCl reacts with 1 mole of calcium carbonate

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$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

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Molar mass of calcium carbonate = 100.01 g/mol

Putting values in the above equation:

$\text{Mass of }CaCO_3=(2.37\times 10^{-4}mol)\times 100.01g/mol\\\\\text{Mass of }CaCO_3=0.0237g$

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