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Tpy6a [65]
3 years ago
5

the reaction below shows how silver chloride can be synthesized. agno3 nacl nano3 agcl how many moles of silver chloride are pro

duced from 15.0 mol of silver nitrate?
Chemistry
1 answer:
Novay_Z [31]3 years ago
8 0
The balanced chemical reaction is:

AgNO3 + NaCl = AgCl + NaNO3

We are given the amount of silver nitrate to be used for the reaction. This will be the starting point of our calculations. 

15.0 mol AgNO3 ( 1 mol AgCl / 1 mol AgNO3 ) = 15.0 mol AgCl
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12.0107

Explanation:

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State the IUPAC name (alcohol)
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Will NaCl be soluble or insoluble
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6 0
4 years ago
You use an analytical balance to measure the mass of a weigh boat and record a mass of 1.5624 g. You then add some NaHCO3 to the
never [62]

Answer:

The minimum number of grams of acetic acid that we would need to fully react with your NaHCO3, is 0.263 grams.

Explanation:

Step 1: Data given

Mass weigh boat = 1.5624 grams

Mass of boat + NaHCO3 = 1.92...

<u>Step 2:</u> Calculate ranges

The minimum possible mass of boat + NaHCO3 = 1.9200g  

Minimum mass of NaHCO3 = 1.9200g - 1.5624g = 0.3576g  

The maximum possible mass of boat + NaHCO3 = 1.9299g  

Maximum mass of NaHCO3 = 1.9299g - 1.5624g = 0.3675g  

What is the minimum mass of acetic acid that must be used to  react with all the NaHCO3 possible in the boat:  

This means what mass of CH3COOH reacts with 0.3675g NaHCO3  

NaHCO3 + CH3COOH → CH3COONa + CO2 + H2O

1 mole of NaHCO3 consumed, needs 1 mole CH3COOH to produce 1 mole of CH3COONa, 1 mole CO2 and 1 mole H2O

<u>Step 3: </u>Calculate moles NaHCO3

Moles NaHCO3 = mass NaHCO3 / Molar mass NaHCO3

moles NaHCO3 = 0.3675 grams / 84g/mol

moles NaHCO3 = 0.004375 moles

Since 1 mole of NaHCO3 consumed, needs 1 mole CH3COOH

For 0.004375 moles NaHCO3, we need 0.004375 moles CH3COOH

<u>Step 4:</u> Calculate mass of CH3COOH

mass of CH3COOH = moles CH3COOH * Molar mass CH3COOH

mass of CH3COOH = 0.004375 moles * 60.05 g/mol

mass CH3COOH = 0.2627 grams ≈ 0.263 grams

The minimum number of grams of acetic acid that we would need to fully react with your NaHCO3, is 0.263 grams.

6 0
3 years ago
What is minimum volume of oxygen required to react with 42.5 g of aluminum in the synthesis of aluminum oxide at stp?
natka813 [3]
<span>26.833 liters

Aluminum oxide has a formula of Al</span>₂O₃,<span> which means for every mole of aluminum used, 1.5 moles of oxygen is required (3/2 = 1.5).

Given 42.5 g of aluminum divided by its atomic mass (26.9815385) gives 1.575 moles of aluminum.

Since it takes 1.5 moles of oxygen per mole of aluminum to make aluminum oxide, you'll need 2.363 moles of oxygen atoms.

Each molecule of oxygen gas has 2 oxygen atoms, so the moles of oxygen gas will be 2.363/2 = 1.1815

Finally, you need to calculate the volume of </span>1.1815 <span>moles of oxygen gas.
1 mole of gas at STP occupies 22.7 liters of volume. Therefore,

1.1815 * 22.7 = </span>26.8 liters <span>of oxygen gas.
</span>
6 0
4 years ago
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