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Vika [28.1K]
4 years ago
10

How many reactant molecules and product gas molecules are in this equation?

Chemistry
1 answer:
Mice21 [21]4 years ago
6 0

Answer:

N₂  = 6.022 × 10²³ molecules

H₂ = 18.066 × 10²³ molecules

NH₃ = 12.044 × 10²³ molecules

Explanation:

Chemical equation;

N₂ + 3H₂     →  2NH₃

It can be seen that there are one mole of nitrogen three mole of hydrogen and two moles of ammonia are present in this equation. The number of molecules of reactant and product would be calculated by using Avogadro number.

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

Number of molecules of nitrogen gas:

1 mol = 6.022 × 10²³ molecules

Number of molecules of hydrogen:

3 mol × 6.022 × 10²³ molecules/ 1 mol

18.066 × 10²³ molecules

Number of molecules of ammonia:

2 mol × 6.022 × 10²³ molecules/ 1 mol

12.044 × 10²³ molecules

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The balanced equation for the above reaction is as follows;
2AgNO₃ + CaCl₂ ---> 2AgCl + Ca(NO₃)₂
stoichiometry of AgNO₃ to CaCl₂ is 2:1
we first need to find the limiting reactant 
number of moles reacted = molarity x volume
number of AgNO₃ moles = 0.22 mol/L x 0.1050 L = 0.023 mol
number of CaCl₂ moles = 0.13 mol/L x 0.1050 L = 0.014 mol
according to molar ratio of 2:1
if we assume AgNO₃ to be the limiting reactant 
if 2 mol of AgNO₃ react with 1 mol of CaCl₂
then 0.023 mol of AgNO₃ reacts with - 0.023/2 = 0.012 mol of CaCl₂
0.012 mol of CaCl₂ is required but 0.014 mol of CaCl₂ is required
therefore CaCl₂ is in excess and AgNO₃ is therefore the limiting reactant 

the amount of products formed depends on the amount of limiting reactant present 
stoichiometry of AgNO₃ to AgCl is 2:2
the number of moles of AgCl formed = number of AgNO₃ moles reacted 
therefore number of AgCl moles formed = 0.023 mol
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mass of AgCl formed = 3.3 g
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Explanation:

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0.0200 M Fe3+ is initially mixed with 1.00 M oxalate ion, C2O42-, and they react according to the equation: Fe3+(aq) + 3 C2O42-(
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Answer : The concentration of Fe^{3+} at equilibrium is 0 M.

Solution :  Given,

Concentration of Fe^{3+} = 0.0200 M

Concentration of C_2O_4^{2-} = 1.00 M

The given equilibrium reaction is,

                            Fe^{3+}(aq)+3C_2O_4^{2-}(aq)\rightleftharpoons [Fe(C_2O_4)_3]^{3-}(aq)

Initially conc.       0.02         1.00                   0

At eqm.             (0.02-x)    (1.00-3x)                x

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K_c=\frac{[[Fe(C_2O_4)_3]^{3-}]}{[C_2O_4^{2-}]^3[Fe^{3+}]}

1.67\times 10^{20}=\frac{(x)^2}{(1.00-3x)^3\times (0.02-x)}

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x=0.02M

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Therefore, the concentration of Fe^{3+} at equilibrium is 0 M.

4 0
3 years ago
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