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Rus_ich [418]
3 years ago
15

The balanced equation for the reaction of aqueous Pb(ClO3)2 with aqueous NaI is Pb(ClO3)2(aq)+2NaI(aq)⟶PbI2(s)+2NaClO3(aq) What

mass of precipitate will form if 1.50 L of concentrated Pb(ClO3)2 is mixed with 0.200 L of 0.210 M NaI? Assume the reaction goes to completion. mass of precipitate:
Chemistry
1 answer:
ryzh [129]3 years ago
6 0

Answer : The mass of PbI_2 precipitate produced will be, 9.681 grams.

Explanation : Given,

Molarity of NaI = 0.210 M

Volume of solution = 0.2 L

Molar mass of PbI_2 = 461.01 g/mole

First we have to calculate the moles of NaI.

\text{Moles of }NaI=\text{Molarity of }NaI\times \text{Volume of solution}=0.210M\times 0.2L=0.042moles

Now we have to calculate the moles of PbI_2.

The balanced chemical reaction is,

Pb(ClO_3)_2(aq)+2NaI(aq)\rightarrow PbI_2(s)+2NaClO_3(aq)

From the balanced reaction we conclude that

As, 2 moles of NaI react to give 1 mole of PbI_2

So, 0.042 moles of NaI react to give \frac{0.042}{2}=0.021 moles of PbI_2

Now we have to calculate the mass of PbI_2.

\text{Mass of }PbI_2=\text{Moles of }PbI_2\times \text{Molar mass of }PbI_2

\text{Mass of }PbI_2=(0.021mole)\times (461.01g/mole)=9.681g

Therefore, the mass of PbI_2 precipitate produced will be, 9.681 grams.

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What is the empirical formula of a compound that contains 6.10 g of hydrogen and 28 g of nitrogen?
Viktor [21]

Answer:

NH₃

Explanation:

mass H = 6.10 grams

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mass cpd = (6.10 + 28.00)grams = 34.10 grams

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%/100wt => grams/100wt => moles => ratio => reduce => emp ratio

%H/100wt = 17.9% w/w => 17.9g => (17.9/1)moles = 17.9 moles H

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Ratio N:H => 17.9 : 5.9

Reduce mole ratio (divide by smaller mole value) => 17.9/5.9 : 5.9/5.9

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3 0
3 years ago
MATCH
almond37 [142]
I wanna say it would be 1 would be A 2 would be C 3 would be D 4 would be B and 5 would be E



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