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3 years ago

What is the balanced equation of iron (iii) hydrogen carbonate?

Chemistry

1 answer:

Lynna [10]3 years ago

6 0

<h3>Answer:</h3>

2Fe(HCO₃)₃ → Fe₂(CO₃)₃ + 3H₂O + 3CO₂

<h3>Explanation:</h3>

A decomposition reaction is one in which a large compound is broken down into smaller compounds or individual elements.
The decomposition reaction Iron (iii) hydrogen carbonate yield iron (iii) carbonate, water and carbon dioxide.

Fe(HCO₃)₃ → Fe₂(CO₃)₃ + H₂O + CO₂

A balanced equation is the one in which the number of atoms of each element are equal on both side of the equation.

Therefore; the balanced equation for the decomposition of Iron (iii) hydrogen carbonate is given by;

2Fe(HCO₃)₃ → Fe₂(CO₃)₃ + 3H₂O + 3CO₂

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3 years ago

Give the chemical symbol of an element in the third period (row) of the periodic table with four 3p electrons.

Ann [662]

Answer: The chemical symbol of the element is Sulfur.

Explanation:

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2 years ago

It is proposed to use Liquid Petroleum Gas (LPG) to fuel spark-ignition engines. A typical sample of the fuel on a volume basis

Norma-Jean [14]

Answer:

The overall balanced combustion reaction is written as :

$0.7C_3H_8 \ + \ 0.05C_4H_{10} \ + \ 0.25 C_3H_6 \ + \ x(O_2 \ + \ 3.76N_2) ---> 3.05CO_2 \ + \ 3.8H_2O \ + \ 18.612N_2$

$(F/A)_{stoichiometric} = 0.0424$

$(A/F)_{stoichiometric} = 23.562$

the higher heating values $(HHV)_f$ per unit mass of LPG = 49.9876 MJ/kg

the lower heating values $(LHV)_f$ per unit mass of LPG = 46.4933 MJ/kg

Explanation:

The stoichiometric equation can be expressed as :

$0.7C_3H_8 \ + \ 0.05C_4H_{10} \ + \ 0.25 C_3H_6 \ + \ x(O_2 \ + \ 3.76N_2) ---> aCO_2 \ + \ bH_2O \ + \ cN_2$

Now, equating the coefficient of carbon; we have:

(0.7×3)+(0.05×4)+(0.25×3) = a

a = 3.05

Also, Equating the coefficient of hydrogen : we have:

(0.7 × 8) +(0.05 × 10) + ( 0.25 × 6) = 2 b

2b = 7.6

b = 3.8

Equating the coefficient of oxygen

2x = 2a + b

$x = \frac{2a+b}{2} \\ \\ x = \frac{2(3.05)+3.8}{2} \\ \\ x = 4.95$

Equating the coefficient of Nitrogen

$c = 3.76x \\ \\ c = 3.76 *4.95 \\ \\ c = 18.612$

Therefore, The overall balanced combustion reaction can now be written as :

$0.7C_3H_8 \ + \ 0.05C_4H_{10} \ + \ 0.25 C_3H_6 \ + \ x(O_2 \ + \ 3.76N_2) ---> 3.05CO_2 \ + \ 3.8H_2O \ + \ 18.612N_2$

Now; To determine the stoichiometric F/A and A/F ratios; we have:

$(F/A)_{stoichiometric} = \frac{n_f}{n_a } \\ \\ (F/A)_{stoichiometric} = \frac{1}{4.95*(1+3.76)} \\ \\ (F/A)_{stoichiometric} = 0.0424$

$(A/F)_{stoichiometric} = \frac{n_a}{n_f } \\ \\ (A/F)_{stoichiometric} = \frac{4.95*(1+3.76)}{1} \\ \\ (A/F)_{stoichiometric} = 23.562$

What are the higher and lower heating values per unit mass of LPG?

Let calculate the molecular mass of the fuel in order to determine their mass fraction of the fuel components.

Molecular mass of the fuel $M_f = (0.7*M_{C_3H_5} ) + (0.05 *M_{C_4H_{10}}) + (0.25*M _{C_3H_6})$

= 30.8 + 2.9 + 10.5

= 44.2 kg/mol

Mass fraction of the fuel components can now be calculated as :

$m_{C_3H_8} = \frac{30.8}{44.2} \\ \\ m_{C_3H_8} = 0.7 \\ \\ \\ m_{C_4H_{10}} = \frac[2.9}{44.2} \\ \\ m_{C_4H_{10}} = 0.06 \\ \\ \\ m_{C_3H_6} = \frac{10.5}{44.2} \\ \\ m_{C_3H_6} = 0.24$

Finally; calculating the higher heating values $(HHV)_f$ per unit mass of LPG; we have:

$(HHV)_f=(0.7 * HHV_{C_3H_8}) + (0.06 *HHV_{C_4H_{10}})+(0.24*HHV_{C_3H_6} \\ \\ (HHV)_f=(0.7*50.38)+(0.06*49.56)+(0.24*48.95) \\ \\ (HHV)_f=49.9876 \ MJ/kg$

calculating the lower heating values $(LHV)_f$ per unit mass of LPG; we have:

$(LHV)_f = (HHV)_f - \delta H_w \\ \\ (LHV)_f = (HHV)_f - [\frac{m_w}{m_f}h_{vap}] \\ \\ (LHV)_f = 49.9876 \ MJ/kg - [\frac{3.8*18}{44.2}*2.258 \ MJ/kg] \\ \\ (LHV)_f = 46.4933 \ M/kg$