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likoan [24]
3 years ago
8

Which set of coefficients will balance this chemical equation?

Chemistry
1 answer:
Montano1993 [528]3 years ago
5 0

Answer:

Option B is correct = 1,3

Explanation:

Chemical equation:

C₂H₄ + O₂     →    CO₂ + H₂O

Balanced chemical equation:

C₂H₄ + 3O₂     →    2CO₂ + 2H₂O

Step 1:

Left side                      Right side

C = 2                           C = 1

H = 4                           O = 3

O = 2                           H = 2

Step 2:

C₂H₄ + O₂     →    2CO₂ + H₂O

Left side                      Right side

C = 2                           C = 2

H = 4                           O = 5

O = 2                           H = 2

Step 3:

C₂H₄ + O₂     →    2CO₂ + 2H₂O

Left side                      Right side

C = 2                           C = 2

H = 4                           O = 6

O = 2                           H = 4

Step 4:

C₂H₄ + 3O₂     →    2CO₂ + 2H₂O

Left side                      Right side

C = 2                           C = 2

H = 4                           O = 6

O = 6                           H = 4

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astra-53 [7]

Given:

Ammonia n h_{3}=6.91 g

Hydrochloric acidh c l=4.61 g

To find:

The amount excess of reactant left over in ammonia.

Solution:

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Equation for the above statement is derived as:

n h_{3}+h c l=n h_{4} c l

One gram per mole of ammonia n h_{3}=\frac{17 g}{m o l e}

Similarly for 6.91g of n h_{3}=\frac{\frac{6.91}{17 g}}{m o l}

                                            =0.40645 \text { moles of } n h_{3}

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Similarly for 4.61g of h c l=\frac{\frac{4.61}{36.45 g}}{\text {mole}}

                                           =0.12647 \text { moles of } h c l

From the above information we can say that h c l is a limiting reactant.

Limiting reactant is an element that consumes lesser product in a chemical reaction.

Thus the amount of excess reactant is calculated by using the following formula

Amount of excess reactant left over in n h_{3}=\text {moles in 6.92 g of } n h_{3}-\text { moles in 4.61 } g \text { of } h c l

                                            =0.40645-0.12647

                                            =0.27998 m o l e s \times \frac{17 g}{m o l e}

                                             =4.75966 g

Result:

Amount of excess reactant left over ammonia n h_{3}=4.75966 g

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Molar mass = Mass / Moles

Upon solving for moles;

Moles = Mass / Molar mass

Moles = 10.8 / 102.09 = 0.1058 mol

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