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brilliants [131]
3 years ago
9

Phosphorus trichloride, PCl3, decomposes to form elemental phosphorus and chlorine. The equation is: 4PCl3 → P4 + 6Cl2. Balance

the equation and determine the mass in grams of chlorine that would be formed if 25 grams of PCl3 (molecular mass = 137.32 g/mol) decompose.
Chemistry
1 answer:
quester [9]3 years ago
8 0
The equation is balanced as written. 
<span>25 g / 137.32 g/mol x (6/4) x 70.9064 g/mol = 19 g of Cl2</span>
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What is the difference between clastic and chemical sedimentary rocks?
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8 0
3 years ago
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Calcium carbonate decomposes when heated to solid calcium oxide and carbon dioxide gas. The balanced equation is: CaCO3(s) → CaO
wlad13 [49]

Answer:

The mass of CaCO3 reacted  is 5.00 grams

Explanation:

Step 1 :Data given

Before the reaction, the container, including the CaCO3, had a mass of 24.20 g

After the reaction the container with product had a mass of only 22.00 g because the CO2 gas produced did not remain in the container.

Molar mass of CO2 = 44.01 g/mol

Molar mass CaCO3 = 100.09 g/mol

Step 2: The balanced equation

CaCO3 → CaO + CO2

Step 3: Calculate mass of CO2

Mass of CO2 = 24.20 grams - 22.00 grams

Mass of CO2 = 2.20 grams

Step 4: Calculate moles CO2

Moles CO2 = mass CO2 / molar mass CO2

Moles CO2 = 2.20 grams / 44.01 g/mol

Moles CO2 = 0.0500 moles

Step 5: Calculate moles CaCO3

For 1 mol CaCO3 we'll have 1 mol CaO and 1 mol CO2

For 0.0500 moles CO2 we need 0.0500 moles CaCO3

Step 6: Calculate mass CaCO3

Mass CaCO3 = moles CaCO3 * molar mass CaCO3

Mass CaCO3 = 0.0500 moles  * 100.09 g/mol

Mass CaCO3 = 5.00 grams

The mass of CaCO3 reacted  is 5.00 grams

8 0
4 years ago
A student observes the chemical process described by this equation:
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Answer:

Both are equal

Explanation:

Mass of starting material i.e Aluminium is 4.480 * 10^{-23} grams

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Thus, the total mass of starting material is

4 * mass of one atom of aluminium + 6 * mass of one atom of oxygen

4 * 4.480 * 10^{-23} +  6* 2.65 * 10^{-23}

35.1 * 10^{-23} grams

Mass of new substance is also equal to

4 * mass of one atom of aluminium + 6 * mass of one atom of oxygen

4 * 4.480 * 10^{-23} +  6* 2.65 * 10^{-23}

35.1 * 10^{-23} grams

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Answer:

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