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3 years ago

Which coefficient before potassium (K) balances this chemical equation?

Chemistry

2 answers:

miskamm [114]3 years ago

5 0

Answer:

The coefficient before potassium (K) balances this chemical equation is 2.

Explanation:

_K +Cl₂ → 2KCl

K =1 ; Cl =2

K=1 × 2 = 2

Cl = 1 × 2 = 2

2 K +Cl₂ = 2 KCl

Zinaida [17]3 years ago

4 0

Answer:

The coefficient before potassium (K) balances this chemical equation is 2.

Explanation:

_K +Cl₂ → 2KCl

K =1 ; Cl =2

K=1 × 2 = 2

Cl = 1 × 2 = 2

2 K +Cl₂ = 2 KCl

Explanation:

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2AlF3 + 3K2O → 6KF + Al2O3

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<h3>Answer:</h3>

7.4797 g AlF₃

<h3>General Formulas and Concepts:</h3>

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<h3>Explanation:</h3>

Step 1: Define

[RxN] 2AlF₃ + 3K₂O → 6KF + Al₂O₃

[Given] 15.524 g KF

Step 2: Identify Conversions

[RxN] 6 mol KF = 2 mol AlF₃

Molar Mass of K - 39.10 g/mol

Molar Mass of F - 19.00 g/mol

Molar Mass of Al - 26.98 g/mol

Molar Mass of KF - 39.10 + 19.00 = 58.1 g/mol

Molar Mass of AlF₃ - 26.98 + 3(19.00) = 83.98 g/mol

Step 3: Stoichiometry

Set up: $\displaystyle 15.524 \ g \ KF(\frac{1 \ mol \ KF}{58.1 \ g \ KF})(\frac{2 \ mol \ AlF_3}{6 \ mol \ KF})(\frac{83.98 \ g \ AlF_3}{1 \ mol \ AlF_3})$
Multiply/Divide: $\displaystyle 7.47966 \ g \ AlF_3$

Step 4: Check

Follow sig fig rules and round. We are given 5 sig figs.

7.47966 g AlF₃ ≈ 7.4797 g AlF₃

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4 years ago

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The smell of dirty gym socks is caused by the compound caproic acid(contains H, C, O). Combustion of 0.844 g of caproic acid pro

adoni [48]

Answer: The molecular formula for the given organic compound is $C_2H_{32}O_4$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=0.784g$

Mass of $H_2O=1.92g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 0.784 g of carbon dioxide, $\frac{12}{44}\times 0.784=0.214g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 1.92 g of water, $\frac{2}{18}\times 1.92=0.213g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.844) - (0.214 + 0.213) = 0.417 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.214g}{12g/mole}=0.0134moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.213g}{1g/mole}=0.213moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.417g}{16g/mole}=0.0261moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0134 moles.

For Carbon = $\frac{0.0134}{0.0134}=1$

For Hydrogen = $\frac{0.213}{0.0134}=15.89\approx 16$

For Oxygen = $\frac{0.0261}{0.0134}=1.95\approx 2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 16 : 2

The empirical formula for the given compound is $CH_{16}O_2$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 116.2 g/mol

Mass of empirical formula = 60 g/mol

Putting values in above equation, we get:

$n=\frac{116.2g/mol}{60g/mol}=1.94\approx 2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 2)}H_{(16\times 2)}O_{(2\times 2)}=C_2H_{32}O_4$

Hence, the molecular formula for the given organic compound is $C_2H_{32}O_4$

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