Chemistry, 16.10.2019 18:30 abc236

Which of the following equations follows the law of conservation of mass?

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Answer from: xxaurorabluexx

The reaction that has been following the law of conservation of mass has been reaction 1. Thus, option A is correct.

The law of conservation of mass has been given as the mass cannot be created nor be destroyed in a chemical reaction. The law of conservation of mass has equal weight of the reactants and the products.

The mass of following reactions can be given as:

$\rm C_2H_4\;+3\;O_2\;\rightarrow\;2\;CO_2\;+2\;H_2O$ 28 g + 3(32)g = 2(44)g + 2(18)g Mass

124 g = 124 g

Thus, the reaction follows the law of conservation of mass.

$\rm C\;+\;4\;H_2\;\rightarrow\;CH_4$

12g + 4(2)g = 16g

20g = 16g

The reactant and product mass has not been equal, thus not follows the law of conservation of mass.

$\rm 3\;H_2O\;\rightarrow\;3\;H_2\;+\;3\;O_2$

3(18) = 3(2) + 3(32)

54g = 102g

The reactant and product mass has not been equal, thus not follows the law of conservation of mass.

$\rm 2\;Na\;+\;Cl\;\rightarrow\;NaCl$

2(23) + 36.5 = 58.5 g

81.5g = 58.5g

The reactant and product mass has not been equal, thus not follows the law of conservation of mass.

The reaction that has been following the law of conservation of mass has been reaction 1. Thus, option A is correct.

For more information about the law of conservation of mass, refer to the link:

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Answer from: gatorr2010

option A = C₂H₄ + 3O₂ → 2CO₂ + 2H₂O

Explanation:

Law of conservation of mass:

This law stated that mass can not be created or destroyed in chemical reaction. It just changed from one to another form.

For example:

C₂H₄ + 3O₂ → 2CO₂ + 2H₂O

28 g + 96 g = 88 g + 36 g

124 g = 124 g

This reaction correctly hold the law of conservation of mass.

Other options:

C + 4H₂ → CH₄

12 g + 8g = 16 g

20 g = 16 g

This reaction do not hold the law of conservation of mass.

3H₂O → 3H₂ + 3O₂

54 g = 6 g + 96 g

54 g = 102 g

This reaction do not hold the law of conservation of mass.

2Na + Cl → NaCl

46 g + 35.5 g = 58.5 g

81.5 g = 58.5 g

This reaction do not hold the law of conservation of mass.

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

Answer from: Quest

I’d have to say no because without oxygen in the water the fish would all be dead. so if you want to fish for dead fish, pick the ponds with the lowest oxygen concentrations.

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