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3 years ago

What is the mass of potassium chloride when 2.50 g of potassium reacts with excess of chlorine gas

Chemistry

1 answer:

Lorico [155]3 years ago

6 0

Answer:

4.52 grams of potassium chloride

Explanation:

First you need a balanced reaction equation K(s) + Cl₂ (g) --> 2 KCl(s). Since the chlorine gas is said to be in excess all of the potassium will be converted to form potassium chloride. Convert grams of potassium to moles of potassium with the periodic table, then you can convert moles of potassium to moles of potassium chloride with the balanced equation, finally convert moles of potassium chloride to grams of potassium chloride with the periodic table. Set up a dimensional analysis chart to help with conversions

(2.50 grams K) * (1 mole K) * (2 mole KCl) * (35.45 grams KCl)

(39.09 grams K) (1 mole K) (1 mol KCl)

Which equals 4.52440778 grams, with sig figs it should be 4.52 grams of potassium chloride

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A certain solid has a density of 8.0 g/cm3. if 4.0 g of this solid are poured into 4.00 ml of water, which drawing below most cl

Alexandra [31]

Answer is: the volume of water after the solid is added is 4.5 ml.
d(gold) = 8.0 g/cm³; density of gold.
m(gold) = 4 g; mass of gold.
V(gold) = m(gold) ÷ d(gold); volume of gold.
V(gold) = 4 g ÷ 8 g/cm³.
V(gold) = 0.5 cm³ = 0.5 ml.
V(water) = 4.00 ml = 4.00 cm³.
V(flask) = V(gold) + V(water).
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3 years ago

When 1.2383 g of an organic iron compound containing Fe, C, H, and O was burned in O2, 2.3162 g of CO2 and 0.66285 g of H2O were

uysha [10]

Answer:

$\boxed{\text{C$_{15}$H$_{21}$FeO$_{6}$}}$

Explanation:

Let's call the unknown compound X.

1. Calculate the mass of each element in 1.23383 g of X.

(a) Mass of C

$\text{Mass of C} = \text{2.3162 g } \text{CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{0.632 07 g C}$

(b) Mass of H

$\text{Mass of H} = \text{0.66285 g }\text{H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g } \text{{H$_{2}$O}}} = \text{0.074 157 g H}$

(c)Mass of Fe

(i)In 0.4131g of X

$\text{Mass of Fe} = \text{0.093 33 g Fe$_{2}$O$_{3}$}\times \dfrac{\text{111.69 g Fe}}{\text{159.69 g g Fe$_{2}$O$_{3}$}} = \text{0.065 277 g Fe}$

(ii) In 1.2383 g of X

$\text{Mass of Fe} = \text{0.065277 g Fe}\times \dfrac{1.2383}{0.4131} = \text{0.195 67 g Fe}$

(d)Mass of O

Mass of O = 1.2383 - 0.632 07 - 0.074 157 - 0.195 67 = 0.336 40 g

2. Calculate the moles of each element

$\text{Moles of C = 0.63207 g C}\times\dfrac{\text{1 mol C}}{\text{12.01 g C }} = \text{0.052 629 mol C}\\\\\text{Moles of H = 0.074157 g H} \times \dfrac{\text{1 mol H}}{\text{1.008 g H}} = \text{0.073 568 mol H}\\\\\text{Moles of Fe = 0.195 67 g Fe} \times \dfrac{\text{1 mol Fe}}{\text{55.845 g Fe}} = \text{0.003 5038 mol Fe}\\\\\text{Moles of O = 0.336 40} \times \dfrac{\text{1 mol O}}{\text{16.00 g O}} = \text{0.021 025 mol O}$

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

$\text{C: } \dfrac{0.052629}{0.003 5038}= 15.021\\\\\text{H: } \dfrac{0.073568}{0.003 5038} = 20.997\\\\\text{Fe: } \dfrac{0.003 5038}{0.003 5038} = 1\\\\\text{O: } \dfrac{0.021025}{0.003 5038} = 6.0006$

4. Round the ratios to the nearest integer

C:H:O:Fe = 15:21:1:6

5. Write the empirical formula

$\text{The empirical formula is } \boxed{\textbf{C$_{15}$H$_{21}$FeO$_{6}$}}$

5 0

3 years ago

If 3.0 moles of hydrogen react, how many grams of nitrogen will be used? Round your answer to the nearest whole number. N2 + 3H2

Iteru [2.4K]

Answer:

28g

Explanation:

Given parameters:

Number of moles of hydrogen = 3moles

Unknown:

Mass of nitrogen used = ?

Solution:

To solve this problem, let us establish the balanced reaction equation first;

N₂ + 3H₂ → 2NH₃

From the balanced reaction equation;

3 mole of H₂ will combine with 1 mole of N₂

Now,

Mass of Nitrogen = number of moles x molar mass

Molar mass of N₂ = 2(14) = 28g/mol

Mass of Nitrogen = 1 mole x 28g/mole = 28g

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The solute has to be hydrophilic, (water loving).

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