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3 years ago

Which one do I pick?

Chemistry

1 answer:

Tomtit [17]3 years ago

4 0

Answer:

i would go for the 2nd as far as i know they were a big continent and an earthquake separes them all

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A student mixes 33.0 mL of 2.70 M Pb ( NO 3 ) 2 ( aq ) with 20.0 mL of 0.00157 M NaI ( aq ) . How many moles of PbI 2 ( s ) prec

GalinKa [24]

Answer: The moles of precipitate (lead (II) iodide) produced is $1.57\times 10^{-5}$ moles

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

For lead (II) nitrate:

Molarity of lead (II) nitrate solution = 2.70 M

Volume of solution = 33.0 mL = 0.033 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$2.70M=\frac{\text{Moles of lead (II) nitrate}}{0.033L}\\\\\text{Moles of lead (II) nitrate}=(2.70mol/L\times 0.0330L)=0.0891mol$

For NaI:

Molarity of NaI solution = 0.00157 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

$0.00157M=\frac{\text{Moles of NaI}}{0.020L}\\\\\text{Moles of NaI}=(0.00157mol/L\times 0.0200L)=3.14\times 10^{-5}mol$

For the given chemical reaction:

$Pb(NO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaNO_3(aq.)$

By Stoichiometry of the reaction:

2 moles of NaI reacts with 1 mole of lead (II) nitrate

So, $3.14\times 10^{-5}$ moles of NaI will react with = $\frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}mol$ of lead (II) nitrate

As, given amount of lead (II) nitrate is more than the required amount. So, it is considered as an excess reagent.

Thus, NaI is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, $3.14\times 10^{-5}$ moles of NaI will produce = $\frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}moles$ of lead (II) iodide

Hence, the moles of precipitate (lead (II) iodide) produced is $1.57\times 10^{-5}$ moles

4 0

3 years ago

What are the two types of stoichiometry?

julsineya [31]

Two types of stoichiometry are; molar mass and coefficients from balanced equation.

8 0

3 years ago

Read 2 more answers

How many milligrams of sodium sulfide are needed to completely react with 25.00 ml of a 0.0100 m aqueous solution of cadmium nit

NARA [144]

Na₂S(aq) + Cd(NO₃)₂(aq) = CdS(s) + 2NaNO₃(aq)

v=25.00 mL
c=0.0100 mmol/mL
M(Na₂S)=78.046 mg/mmol

n(Na₂S)=n{Cd(NO₃)₂}=cv

m(Na₂S)=M(Na₂S)n(Na₂S)=M(Na₂S)cv

m(Na₂S)=78.046*0.0100*25.00≈19.5 mg

5 0

4 years ago

Read 2 more answers

2. Which equation represents a transmutation?

antiseptic1488 [7]

8935982560 plz contact me

5 0

3 years ago

One way of obtaining pure sodium carbonate is through the decomposition of the mineral trona, Na3(CO3)(HCO3)·2H2O. 2Na3(CO3)(HCO

zhenek [66]

Percentage yield = (actual yield / theoretical yield) x 100%

The balanced equation for the decomposition is,
2Na₃(CO₃)(HCO₃)·2H₂O(s) → 3Na₂CO₃(s) + CO₂(g) + 5H₂O(g)

The stoichiometric ratio between Na₃(CO₃)(HCO₃)·2H₂O(s) and Na₂CO₃(s) is 2 : 3

The decomposed mass of Na₃(CO₃)(HCO₃)·2H₂O(s) = 1000 kg
= 1000 x 10³ g

Molar mass of Na₃(CO₃)(HCO₃)·2H₂O(s) = 226 g mol⁻¹
moles of Na₃(CO₃)(HCO₃)·2H₂O(s) = mass / molar mass
= 1000 x 10³ g / 226 g mol⁻¹
= 4424.78 mol

Hence, moles of Na₂CO₃ formed = 4424.78 mol x $\frac{3}{2}$
= 6637.17 mol

Molar mass of Na₂CO₃ = 106 g mol⁻¹

Hence, mass of Na₂CO₃ = 6637.17 mol x 106 g mol⁻¹
= 703540.02 g
= 703.540 kg

Hence, the theoretical yield of Na₂CO₃ = 703.540 kg
Actual yield of Na₂CO₃ = 650 kg

Percentage yield = (650 kg / 703.540 kg) x 100%
= 92.34%

7 0

3 years ago