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4 years ago

Write the MOLECULAR EQUATION between zinc and hydrochloric acid.

Chemistry

1 answer:

jenyasd209 [6]4 years ago

8 0

Answer: The molecular equation is written below.

Explanation:

Molecular equation is defined as the equation in which ionic compounds are expressed as compounds instead of constituent ions.

The balanced molecular equation for the reaction of zinc and hydrochloric acid follows:

$Zn(s)+2HCl(aq.)\rightarrow ZnCl_2(aq.)+H_2(g)$

By Stoichiometry of the reaction:

1 mole of zinc metal reacts with 2 moles of hydrochloric acid to produce 1 mole of zinc chloride and 1 mole of hydrogen gas.

Hence, the molecular equation is written above.

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Answer:

63.25 grams of CO₂

Explanation:

To convert from liters to grams, we first need to convert from liters to moles. To do this, we divide the liters by 22.4, the amount of liters of a gas per mole.

32.2 / 22.4

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This is your answer.

Hope this helps!

7 0

3 years ago

Consider the balanced equation for the following reaction:

Zanzabum

Answer: The amount of carbon dioxide formed in the reaction is 5.663 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of oxygen gas = 8 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{8g}{32g/mol}=0.25mol$

For the given chemical equation:

$7O_2(g)+2C_2H_6(g)\rightarrow 4CO_2(g)+6H_2O(l)$

By Stoichiometry of the reaction:

7 moles of oxygen gas produces 4 moles of carbon dioxide

So, 0.25 moles of oxygen gas will produce = $\frac{4}{7}\times 0.25=0.143mol$ of carbon dioxide

Now, calculating the mass of carbon dioxide from equation 1, we get:

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = 0.143 moles

Putting values in equation 1, we get:

$0.143mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=(0.143mol\times 44g/mol)=6.292g$

To calculate the experimental yield of carbon dioxide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Percentage yield of carbon dioxide = 90 %

Theoretical yield of carbon dioxide = 6.292 g

Putting values in above equation, we get:

$90=\frac{\text{Experimental yield of carbon dioxide}}{6.292g}\times 100\\\\\text{Experimental yield of carbon dioxide}=\frac{90\times 6.292}{100}=5.663g$

Hence, the amount of carbon dioxide formed in the reaction is 5.663 grams

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