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attashe74 [19]
3 years ago
14

Write a net ionic equation for the reaction of Iron(III) Chloride and Potassium Thiocyanate.

Chemistry
1 answer:
zysi [14]3 years ago
3 0

FeCl3 + 6KSCN -------> K3[Fe(SCN)6] + 3KCl


Fe³⁺ + 3Cl⁻ +6K⁺ +6SCN⁻ -----> 3K⁺ + [Fe(SCN)6]³⁻ + 3K⁺ +3Cl⁻


A net ionic equation:


Fe³⁺ + 6SCN⁻ --------> [Fe(SCN)6]³⁻


You probably need this equation in the different form

FeCl3(aq) + 3KSCN(aq) -------> Fe(SCN)3(s) + 3KCl(aq)

Then net equation:

Fe³⁺(aq) + 3SCN⁻(aq) ----->  Fe(SCN)3(s)


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Answer:

Explanation:

1. find the molar mass (amu) of each element and add them to get the whole molar mass.

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<h2><u><em>56-57: NaCl</em></u></h2>

1. Na(22.99amu) + Cl (35.453amu)=58.443

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<h2><u>58-60 </u>K_{2} CO_{3}<u /></h2>

1. K: (39.098)(2)=78.196

_ C: (12.011)(1)= 12.011

_O: (15.99)(3) = 47.997

78.196+12.011+47.997= 138.204

2:K: \frac{78.196}{138.204}= .566 <u>Step </u>3: (.566)(100)= 56.6%

2: C: \frac{12.011}{138.204}= .087 <u>Step 3</u>: (.087)(100)= 8.7%

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<h2>61-62 Fe_{3} O_{4}</h2>

1. Fe (55.845)(3)= 167.535

_ O (15.999)(4) = 63.996

167.535+63.996=231.531

2: Fe: \frac{167.535}{231.531}= .724 Step 3: (.724)(100)= 72.4%

2: O : \frac{63.996}{231.531}= .276 Step 3: (.276)(100) = 27.6%

<h2>63-65 C_{3}H_{5}(OH)_{3}</h2>

1.

C(12.011*3)=36.033

H(1.008*5)=5.04 + (1.008*3)=3.024 so its 8.064

O(15.999*3)=47.997

add them: 92.094

2: C: \frac{36.033}{92.094}= .391 Step 3: (.391)(100) = 39.1%

2: H: \frac{8.064}{92.094}= .088 step 3: (.088)(100) = 8.8%

2: O: \frac{47.997}{92.094} = .521 step 3: (.521)(100) = 52.1%

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Explanation :

According to the Boyle's, law, the pressure of the gas is inversely proportional to the volume of gas at constant temperature and moles of gas.

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V_1 = initial volume = 500.0 L

V_2 = final volume at STP = ?

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2atm\times 500.0L=1atm\times V_2

V_2=1000L

Therefore, the final volume at STP is, 1000 L

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