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Ede4ka [16]
3 years ago
14

-FeS2+-O2->-Fe2O3+-SO2 What coefficient should be placed in front of SO2 to balance the equation?

Chemistry
1 answer:
NISA [10]3 years ago
8 0

Answer:

8 should be placed in front of SO₂.

Explanation:

The easiest way to solve this question is by writing the <u>entire</u> balanced equation:

4FeS₂ + 11O₂ -> 2Fe₂O₃ + 8SO₂

We can achieve this by first balancing the Fe, then S, and finally the O.

We can also double check our answer by counting the number of each element on both sides:

-Reactants: 4 Fe, 8 S, 22 O

-Products: 4 Fe 8 S, 22 O

Since they match, our equation is balanced, and the coefficient in front of SO₂ is 8.

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Protons and Neutrons

Explanation:

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3 years ago
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Is the following compound an alcohol, aldehyde, or organic acid?<br> C-C=0
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6 0
3 years ago
This is a quiz and it’s due today
Romashka [77]
The answer is the el second one since it is a physical change
7 0
3 years ago
Determine the enthalpy change for the decomposition of calcium carbonate. CaCO₃(s) --&gt; CaO(s) + CO₂(g) given the thermochemic
Hunter-Best [27]

Answer: c. 179 kJ/mol

Explanation:

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to Hess’s law, the chemical equation can be treated as algebraic expressions and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

Given:

Ca(OH)_2(s)\rightarrow CaO(s)+H_2O (l)   \Delta H_1= 65.2 kJ/mol     (1)

Ca(OH)_2(s)+CO_2(g)\rightarrow CaCO_3(s)+H_2O(l)    \Delta H_2= -113.8 kJ/mol   (2)

C(s)+O_2(g)\rightarrow CO_2(g)    \Delta H_3= -393.5 kJ/mol   (3)

2Ca(s)+O_2(g)\rightarrow 2CaO(s) \Delta H_4=-1270.2 kJ/mol    (4)

On subtracting eq (1) from eq (2) we have:

Ca(OH)_2(s)\rightarrow CaO(s)+H_2O(l) - Ca(OH)_2(s)+CO_2(g)\rightarrow CaCO_3(s)+H_2O(l)

CaCO_3(s)\rightarrow CaO(s)+CO_2(g)

\Delta H=-113.8-(65.2)kJ/mol=-179kJ/mol

Hence the enthalpy change for the raection is 179.0 kJ/mol.

6 0
3 years ago
When zinc oxide reacts with dilute nitric acid, zinc nitrate is produced. The equation for the reaction is: ZnO + 2HNO₃→ Zn(NO₃)
Lisa [10]

ZnO + 2HNO₃→ Zn(NO₃)₂ + H₂O

mol Zn(NO₃)₂ = mol ZnO = 0.2 (from the coefficient)

mass Zn(NO₃)₂ = 0.2 x 189 = 37.8 g (theorethical yield)

% yield = 15.3 : 37.8 x 100% = 40.48%

6 0
2 years ago
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