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puteri [66]
3 years ago
15

How do I do a balancing equation. I understand. Can you show me a step by step on how to do one of the equations that aren't don

e?

Chemistry
1 answer:
sasho [114]3 years ago
8 0
To balance a chemical equation, you first must count the number of each of the atoms on both the left hand side - reactant side and the right hand side - product side of the chemical reaction.

After counting the atoms, make a tally of the atoms on both sides, for instance if it is hydrogen - h, record the number of hydrogen atoms on the reactant and product side. Then, place coefficients at appropriate places for the elements and or compounds, to basically have all of the different kinds of atoms from their respective elements equal the same on both the left and right side of the reaction.

SnO2 + H2 > Sn + H2O
1st count the metals - Sn atoms on both sides -
Sn left side - 1
Sn right side - 1

2nd count the oxygen atoms on both sides
O atom left side - 2
O atom right side - 1.

Here, the atoms are not the same, so place a coefficient on the side with the least number of oxygen atoms, right side, put 2 to balance.

Now O atoms are same on both sides, 2 atoms but, there are 4 hydrogen atoms now,

Since 2H2O = 2H2 + 2O = H = 4 and O = 2.

Now place the coefficient of 2 on the side with Hydrogen atoms by itself, left side, now this also gives us 4 hydrogens.

So balanced equation:
SnO2 + 2H2 > Sn + 2H2O.
You might be interested in
39. Analyze What subscripts would you most likely use if
Igoryamba

Based on their valencies, the subscripts of the ionic compounds formed will be:

  • 1 and 1
  • 2 and 1
  • 1 and 2
  • 2 and 2

<h3>What subscripts would you most likely use if the following substances formed an ionic compound?</h3>

A. An alkali metal and a halogen.

An alkali metal and a halogen both have valencies of one.

Therefore the subscripts would be 1.

B. An alkali metal and a non-metal from group 16.

An alkali metal has a valency of 1 and a group 16 non-metal has a valency of 2.

By exchange of valencies, the subscript would be 2 and 1.

C. An alkaline earth metal and a halogen.

An alkaline earth metal has a valency of 2 and a halogen has a valency of 1.

By exchange of valencies, the subscripts would be 1 and 2.

D. An alkaline earth metal and a non-metal from group 16.

An alkaline earth metal has a valency of 2 and a non-metal from group 16 has a valency of 2.

By exchange of valencies, the subscripts would be 2 and 2.

Therefore, the subscripts of the ionic compounds formed will be:

  • 1 and 1
  • 2 and 1
  • 1 and 2
  • 2 and 2

Learn more about about valency at: brainly.com/question/2284519

4 0
2 years ago
Explain why gasoline is insoluble in water
SVETLANKA909090 [29]

Answer:

it's not

Explanation:

gasoline is not soluble in water in gasoline is a complex mixture of nonpolar compounds such as long-chain hydrocarbons etc water is a polar molecule as a result of the gasoline molecules cannot enter water and borla solution

4 0
3 years ago
A student has a sample of 1.33 moles of fluorine gas that is contained in a 23.3 L container at 314 K. What is the pressure of t
Gelneren [198K]

1.4715 atm is the pressure of the sample 1.33 moles of fluorine gas that is contained in a 23.3 L container at 314 K.

What is an ideal equation?

The ideal gas equation, pV = nRT, is an equation used to calculate either the pressure, volume, temperature or number of moles of a gas. The terms are: p = pressure, in pascals (Pa).

Given data:

Volume (V) = 23.3 L

Number of mole (n) = 1.33 moles

Temperature (T) = 314 K

Gas constant (R) = 0.821 atm.L/Kmol

Pressure (P) =?

The pressure inside the container can be obtained by using the ideal gas equation as illustrated below:

PV = nRT

P × 23.3 L = 1.33 moles × 0.0821 ×314 K

P = 1.4715 atm

Therefore, the pressure of the sample is 1.4715 atm.

Learn more about the ideal gas equation:

brainly.com/question/23826793

SPJ1

4 0
2 years ago
PLEASE HELPPP ASAPPPPP!!!!!
lutik1710 [3]
Most acids are liquids!! Hope that helps
6 0
3 years ago
Escriba en termino de moles, de moléculas y de masa las siguientes ecuaciones a. Fe +2HCl ________ FeCl2 + H2 b. CH4 + 2O2 _____
MA_775_DIABLO [31]

Answer:

a.

  • 1 mol de Hierro reacciona con 2 moles de acido clorhidrico para formar 1 mol de cloruro de hierro (II)
  • 6.02×10²³ moleculas de hierro reaccionan con 1.20×10²⁴ moleculas de acido clorhidrico para formar 6.02×10²³ moleculas de cloruro de hierro (II)
  • 55.85 g de hierro reaccionan con 72.9 gramos de acido clorhidrico para formar 126.75 g de cloruro férrico.

b.

  • 6.02×10²³ moleculas de metano reaccionan con 1.20×10²⁴ moleculas de oxigeno para formar 6.02×10²³ moleculas de dioxido de carbono y 1.20×10²⁴ moleculas de agua.
  • 1 mol de metano reacciona con dos moles de oxigeno para generar 1 mol de dióxido de carbono y dos moles de agua en estado de vapor.
  • 16 gramos de metano reaccionan con 64 g de oxigeno para formar 44 gramos de dioxido de carbono y 36 gramos de agua.

c.

  • 3 moles de plata sólida reaccionan con 4 moles de acido nitrico para formar 3 moles de nitrato de plata, 1 mol de monoxido de nitrogeno y 2 moles de agua.
  • 1.80×10²⁴ moleculas de plata reaccionan con 2.41×10²⁴ moleculas de acido nitrico para formar 1.80×10²⁴ moleculas de nitrato de plata, 6.02×10²³ moleculas de monoxido de nitrogeno y 1.20×10²⁴ moleculas de agua.
  • 323.58 g de plata reaccionan con 252 g de acido nitrico para formar 509.58 g de nitrato de plata, 30 g de monoxido de nitrogeno y 36 g de agua.

Explanation:

a. Fe (s)  +2 HCl (aq) → FeCl₂ (aq)

1 mol de Hierro reacciona con 2 moles de acido clorhidrico para formar 1 mol de cloruro de hierro (II)

Calculamos cuanto son dos moles de moleculas sabiendo que:

6.02×10²³ moleculas / 1 mol  . 2 mol = 1.20×10²⁴ moleculas. Entonces

6.02×10²³ moleculas de hierro reaccionan con 1.20×10²⁴ moleculas de acido clorhidrico para formar 6.02×10²³ moleculas de cloruro de hierro (II)

Calculamos las masas molares de cada reactivo y producto

Fe = 55.85 g

HCl = 36.45 g

FeCl₂ = 126.75 g

55.85 g de hierro reaccionan con 72.9 gramos de acido clorhidrico para formar 126.75 g de cloruro férrico.

b. CH₄(g) + 2O₂ (g) → CO₂ (g) + 2H₂O (g)

1 mol de metano reacciona con dos moles de oxigeno para generar 1 mol de dióxido de carbono y dos moles de agua en estado de vapor.

6.02×10²³ moleculas de metano reaccionan con 1.20×10²⁴ moleculas de oxigeno para formar 6.02×10²³ moleculas de dioxido de carbono y 1.20×10²⁴ moleculas de agua.

Calculamos las masas molares:

CH₄ = 16 g

O₂ = 32 g

CO₂ = 44 g

H₂O g = 18 g

16 gramos de metano reaccionan con 64 g de oxigeno para formar 44 gramos de dioxido de carbono y 36 gramos de agua.

c. 3 Ag (s) + 4HNO3 (aq) → 3 AgNO3 (aq) + NO (g) + 2H₂O (aq)

Calculamos cuantos moleculas contienen 3 y 4 moles:

6.02×10²³  . 3 = 1.80×10²⁴ moleculas

6.02×10²³  . 4 = 2.41×10²⁴ moleculas

3 moles de plata sólida reaccionan con 4 moles de acido nitrico para formar 3 moles de nitrato de plata, 1 mol de monoxido de nitrogeno y 2 moles de agua.

1.80×10²⁴ moleculas de plata reaccionan con 2.41×10²⁴ moleculas de acido nitrico para formar 1.80×10²⁴ moleculas de nitrato de plata, 6.02×10²³ moleculas de monoxido de nitrogeno y 1.20×10²⁴ moleculas de agua.

Calcualmos las masas molares:

Ag = 107.86 g

HNO₃ = 63 g

AgNO₃ = 169.86 g

NO = 30 g

H₂O = 18 g

323.58 g de plata reaccionan con 252 g de acido nitrico para formar 509.58 g de nitrato de plata, 30 g de monoxido de nitrogeno y 36 g de agua.

6 0
3 years ago
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