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3 years ago

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If 4.52g of the precipitate is formed, how many grams of sodium chloride reacted?

1 answer:

s2008m [1.1K]3 years ago

7 0

The question is incomplete, the complete question is;

AgNO3(aq)+NaCl(aq)→AgCl(s)⏐↓+NaNO3(aq)

1- If 4.52g of the precipitate is formed, how many grams of sodium chloride reacted?

Answer:

1.8 g

Explanation:

Given the equation;

AgNO3(aq)+NaCl(aq)→AgCl(s)↓+NaNO3(aq)

We can see that the reaction is 1:1 Hence, 1 mole of sodium chloride yielded 1 mole of the precipitate(AgCl).

If this is so,

Number of moles of precipitate formed = 4.52g/143.32 g/mol

Number of moles of precipitate formed = 0.0315 moles

Hence, 0.0315 moles of precipitate was formed by 0.0315 moles of NaCl

Therefore;

Mass of NaCl reacted = 0.0315 moles * 58.5 g/mol = 1.8 g

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Read 2 more answers

How many mL will a 0.205 mole sample of He occupy at 3.00 atm and 200 K? Report your answer to the nearest mL.

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1.1214 mL will a 0.205-mole sample of He occupy at 3.00 atm and 200 K.

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

Using equation PV=nRT, where n is the moles and R is the gas constant. Then divide the given mass by the number of moles to get molar mass.

Given data:

P= 3.00 atm

V= ?

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R= $0.082057338 \;L \;atm \;K^{-1}mol^{-1}$

T=200 K

Putting value in the given equation:

$\frac{nRT}{P} =V$

$V= \frac{0.205 \;mole\;0.082057338 \;L \;atm \;K^{-1}mol^{-1} X 200}{3 \;atm}$

V= 1.1214 mL

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