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4 years ago

How many grams of silver chloride are produced from 5.0g of silver nitrate reacting with an excess of

Chemistry

1 answer:

Sav [38]4 years ago

6 0

Answer:

4.21 g of AgCl

3.06 g of BaCl₂ will be needed to complete the reaction

Explanation:

The first step is to determine the reaction.

Reactants: BaCl₂ and AgNO₃

The products will be the silver chloride (AgCl) and the Ba(NO₃)₂

The reaction is: BaCl₂(aq) + 2AgNO₃(aq) → 2AgCl(s) ↓ + Ba(NO₃)₂ (aq)

We determine the silver nitrate moles: 5 g . 1mol / 169.87 g = 0.0294 moles. Now, according to stoichiometry, we know that ratio is 2:2-

2 moles of nitrate can produce 2 moles of chloride, so the 0.0294 moles of silver nitrate, will produce the same amount of chloride.

We convert the moles to mass → 143.32 g / mol . 0.0294 mol = 4.21 g of AgCl.

Now, we consider the BaCl₂.

2 moles of nitrate can react to 1 mol of barium chloride

Then, 0.0294 moles of silver nitrate will react to (0.0294 . 1) /2 = 0.0147 moles. We convert the moles to mass:

0.0147 mol . 208.23 g /1mol = 3.06 g of BaCl₂

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2 years ago

3. The molar volume of a gas at STP occupies 0 1 kilopascal O ooc 12 grams 22.4 L

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Option D

Explanation:

To find the volume of 1 mole of a gas at STP, we use the Ideal Gas Law. It is the general gas equation which gives the relation to the measurable quantities to an ideal gas as below,

P (pressure) × V (volume) = n (number of moles) × R (the gas constant) × T (temperature in Kelvin)

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P = 1 atm

V = ?

n = 1 mole

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T = 273 K

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