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Likurg_2 [28]
4 years ago
13

Write a balance equation for the following.

Chemistry
1 answer:
fomenos4 years ago
6 0
First you must determing the formulas of both of the compounds

Tin (II) Chloride

the (II) means that Tin is 2+ in this compounds; and we know that Cl is -
So the formula for Tin (II) Chloride has to be

SnCl₂

For Sodium Sulfide we know that Sodium is + and Sulfide Ions are 2- so the formula for Sodium Sulfide has to be

Na₂S

As of right now our equation is

SnCl₂ + Na₂S →

When these two compounds react they will form Sodium Chloride and Tin Sulfide

The equations for the two products will be

NaCl      and SnS

now our equation looks like this

SnCl₂ (aq) + Na₂S (aq) → NaCl + SnS

for this to be complete however, we need to first add phases to each of the products and balance the equation

We know that all Group one elements are soluble in water so this means NaCl is soluble


SnCl₂ (aq) + Na₂S (aq) → NaCl (aq) + SnS

we also know that all sulfides are insoluble which means SnS is insoluble

SnCl₂ (aq) + Na₂S (aq) → NaCl (aq) + SnS (s)

Now that we have the phases written in we must balance the equation
There is 1 Sn, 2Cl, 2 Na and 1 S on the left/reactants side
 and 1 Na, 1 Cl, 1Sn, and 1S on the right/products side

to balance it we need to have 2 Cl on the right hand or products side

SnCl₂ (aq) + Na₂S (aq) → 2 NaCl (aq) + SnS

Now everything is balanced!

SnCl₂ (aq) + Na₂S (aq) → 2 NaCl (aq) + SnS is our final reaction equation


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The atomic mass of sulfur is 32.1 amu, and the atomic mass of oxygen is 16.0 amu. To the nearest tenth of a percent, what is the
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Answer:

\%\ Composition\ of\ sulfur=40.1\ \%

Explanation:

Percent composition is percentage by the mass of element present in the compound.

Given , Mass of sulfur= 32.1 amu

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Mass of sulfur trioxide SO_3 = 32.1 amu + 3*16.0 amu = 80.1 amu

\%\ Composition\ of\ sulfur=\frac{Mass_{sulfur}}{Mass_{SO_3}}\times 100

\%\ Composition\ of\ sulfur=\frac{32.1\ amu}{80.1\ amu}\times 100=40.1\ \%

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