The correct name for HIO2?

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

2 years ago

How many moles of aluminum are needed to react completely with 1.2 mol of feo

Amanda [17]

Answer:

0.8 mol.

Explanation:

The balanced equation for the reaction between Al and FeO is represented as:

2Al + 3FeO → 3Fe + Al₂O₃,

It is clear that 2 mol of Al react with 3 mol of FeO to produce 3 mol of Fe and 1 mol of Al₂O₃.

Using cross multiplication:

2 mol of Al needs → 3 mol of FeO, from stichiometry.

??? mol of Al needs → 1.2 mol of FeO.

∴ The no. of moles of Al are needed to react completely with 1.2 mol of FeO = (2 mol)(1.2 mol)/(3 mol) = 0.8 mol.

3 0

3 years ago

The essential oil found in cloves, eugenol, can be isolated by steam distillation because it is insoluble in water and has a mea

Damm [24]

We know that:
Molar Mass H2O: 18 g/mol
Molar Mass of Eugenol: 164 g/mol 
Boiling point of H2O: 100 degrees C 
Boiling point of Eugenol: 254 degrees C 
Density of water: 1.0 g/mL 
Density of Eugenol: 1.05 g/mL 

Using formula:
V= [mole fraction x molar mass] / density 

mH20: 0.9947 * 18
= 17.9046 / 1 g/mL
= 17.9046 
morg: 0.0053 * 164
= 0.8692/ 1.05 g/mL
= 0.8278 

V% = Vorg/(Vorg + VH2O) * 100 
(0.8278/18.7324) * 100 = 4.419% 

Yotal volume = 30 mL; therefore,
0.0442 = (volume eugenol/30) 

(m eug/mH2O) = (peug*164/pH2O*18) 
(m eug/30) = (4*164/760*18) 
m eug = about 1.44g and 

volume = mass/density
= 1.44/1.05
= about 1.37 mL

6 0

2 years ago

Why is hydgron content with only 2 electrons instead of 8 like atoms in periods greater than 1?

Cerrena [4.2K]

Because the sublevels 1s only has one shape ,and it only contains 2 elections...

5 0

2 years ago

Which statement correctly describes an endothermic chemical reaction?

Natasha2012 [34]

An endothermic chemical reactions products tend to have a higher potential energy than the reactants. Thus making the /\ H positive, not negative.

7 0

3 years ago