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1 year ago

Nitrogen and hydrogen combine to form ammonia.

Chemistry

1 answer:

zhenek [66]1 year ago

6 0

The balanced chemical equation for the formation ammonia is

N2(g) + 3H2(g) ----> 2NH3(g) .

The balanced chemical equations explains that the same number of each element exist as reactants and products. The coefficients in a balanced equation must be the simplest whole number ratio. Mass is always conserved in chemical reactions.

For the formation of ammonia, the chemical equation is

N2(g) + H2(g) ----> NH3(g)

Balancing the chemical reaction, we can write,

N2(g) + 3H2(g) ----> 2NH3(g) .

This equation shows two nitrogen entering the reaction together and two hydrogens entering the reaction together. Since NH3 is multiplied by a coefficient of 2 there are now 2 nitrogen and 6 hydrogens. The 6 hydrogens come from the 2 multiplied by the subscript of 3. This is the balanced chemical reaction.

To learn more about Balanced chemical equation please visit:

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6. Write the ICE chart for the reaction of 32.0 g of sulfur and 71.0 g of chlorine: S8 + 4 Cl24S2Cl2 After completing the chart

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Answer:

ICE Table Figure

a. 67.37 g $S_2Cl_2$

b. 35.62 g $Cl_2$

c. 58.61 g $S_2Cl_2$

Explanation:

For the ICE table we have to keep in mind that we have 4 moles of $Cl_2$ and 1 mol of $S_8$ and the reactives are consumed, so for $Cl_2$ we will have -4X and for $S_8$ we will have -X. Follow the same logic we will have -4X for $S_2Cl_2$ .

a. Mass of the product

Molar mass of $S_8$ = 256.52 g/mol

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Molar mass of $S_2Cl_2$ =135.03 g/mol

We have to find the limiting reagent in the reaction:

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$\frac{32}{256.52}=0.124molS_8$

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Divide by the coefficients in the balanced reaction:

$\frac{0.124}{1}=0.124mol$

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The limiting reagent would be $S_8$

Now is posible to calculate the amount of $S_2Cl_2$ produced:

$0.124molS_8\frac{4molS_2Cl_2}{1 molS_8}\frac{135.03gS_2Cl_2}{1 molS_2Cl_2}=67.37gS_2Cl_2$

b. Mass in excess

$0.124molS_8\frac{4molCl_2}{1 molS_8}\frac{70.9gCl_2}{1 molCl_2}=35.38gCl_2$

$Excess\hspace{0.1cm}=\hspace{0.1cm}71gCl_2-35.38gCl_2=35.62 gCl_2$

C. 87%Yield

$67.37gS_2Cl_2\frac{87}{100}=58.61gS_2Cl_2$

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