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pychu
2 months ago
14

What is the net ionic equation of the reaction of mgso4 with sr(no3)2? express you answer as a chemical equation including phase

s?
Chemistry
2 answers:
castortr0y [3K]2 months ago
6 0
MgSO4 and Sr(NO3)2 are soluble, leading to the following ionic equation: Mg+2 + SO4-2 + Sr+2 + 2NO3-1 → Mg+2 + 2NO3-2 + SrSO4. Since SrSO4 is insoluble, we cancel out Mg and 2NO3 from both sides, resulting in the net ionic equation: SO4-2 + Sr+2 → SrSO4
lions [2.9K]2 months ago
6 0
The net ionic equation for the reaction involving MgSO4 and Sr(NO3)2 is:  

Sr⁺²(aq) + SO₄⁻²(aq) → SrSO₄(s)

Additional Explanation

Net ionic equations

  • exclude certain ions from reactants and products. The equations focus solely on the ions that engage in the reaction. Spectator ions, which do not participate, are omitted from the ionic equations.
  • The reaction between MgSO4 and Sr(NO3)2 exemplifies a displacement process.
  • During this interaction, magnesium ions react with nitrate ions to generate magnesium nitrate, while strontium ions react with sulfate ions to form strontium sulfate.

This is represented as:

MgSO₄(aq) + Sr(NO₃)₂(aq) → Mg(NO₃)₂(aq) + SrSO₄(s)

  • The magnesium ion remains aqueous in both magnesium nitrate and magnesium sulfate, thus it gets canceled as a spectator ion.
  • Similarly, the nitrate ion also acts as a spectator ion since it doesn't change state and is eliminated from the net ionic equation.

Consequently, the needed ionic equation will be:

Sr⁺²(aq) + SO₄⁻²(aq) → SrSO₄(s)

Key Terms: Net ionic equations

Learn more about;

  • Net ionic equation:
  • Spectator ions:
  • Example of net ionic equations;

Level: High school

Subject: Chemistry

Topic: Chemical equation

Sub-topic: Net ionic equations
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Answer: The enthalpy change for the reaction is, 201.9 kJ

Explanation:

Based on Hess’s law of constant heat summation, the energy released or absorbed in a chemical reaction stays constant, regardless of whether the process unfolds in one step or multiple steps.

This principle implies, that chemical equations can be treated analogously to algebraic expressions, allowing addition or subtraction to create the needed equation. Thus, the overall enthalpy change corresponds to the summation of the individual enthalpy changes of the reactions occurring in between.

The balanced equation for CH_4 appears as follows,

2CH_4(g)\rightarrow C_2H_4(g)+2H_2(g)    \Delta H^o=?

The intermediate balanced reactions are outlined as follows,

(1) CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)     \Delta H_1=-890.3kJ

(2) C_2H_4(g)+H_2(g)\rightarrow C_2H_6(g)     \Delta H_2=-136.3kJ

(3) 2H_2(g)+O_2(g)\rightarrow 2H_2O(l)    \Delta H_3=-571.6kJ

(4) 2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(l)     \Delta H_4=-3120.8kJ

Next, we will multiply the first reaction by 2, reverse the second, and reverse and halve the third and fourth reactions before combining them. This gives us:

(1) 2CH_4(g)+4O_2(g)\rightarrow 2CO_2(g)+4H_2O(l)     \Delta H_1=2\times (-890.3kJ)=-1780.6kJ

(2) C_2H_6(g)\rightarrow C_2H_4(g)+H_2(g)    \Delta H_2=-(-136.3kJ)=136.3kJ

(3) H_2O(l)\rightarrow H_2(g)+\frac{1}{2}O_2(g)    \Delta H_3=-\frac{1}{2}\times (-571.6kJ)=285.8kJ

(4) 2CO_2(g)+3H_2O(l)\rightarrow C_2H_6(g)+\frac{7}{2}O_2(g)     \Delta H_4=-\frac{1}{2}\times (-3120.8kJ)=1560.4kJ

Therefore, the expression for the enthalpy of the reaction is,

\Delta H^o=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4

\Delta H=(-1780.6kJ)+(136.3kJ)+(285.8kJ)+(1560.4kJ)

\Delta H=201.9kJ

Hence, the enthalpy change for this reaction is, 201.9 kJ

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