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3 years ago

find the percent yield,if 6.0g of Licl are actually produced and the theoretical yield is 35.4g of Licl.

Chemistry

1 answer:

drek231 [11]3 years ago

8 0

Answer:

Percent yield = 17%

Explanation:

Given data:

Actual yield of lithium chloride = 6.0 g

Theoretical yield of lithium chloride = 35.4 g

Percent yield = ?

Solution:

Formula:

Percent yield = (actual yield / theoretical yield )× 100

Now we will put the values in formula.

Percent yield = (6.0 g/ 35.4 g)× 100

Percent yield = 0.17 × 100

Percent yield = 17%

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marishachu [46]

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3 years ago

Read 2 more answers

Determine the total number of atoms contain in a 2.00 moles of Ni

spayn [35]

Answer:

2.00 moles of Ni has 1.2 *10^24 atoms

Explanation:

Step 1: Data given

Number of moles Ni = 2.00 moles

Number of Avogadro = 6.022*10^23 /mol

Step 2: Calculate number of atoms

Number of particles (=atoms) = Number of Avogadro * number of moles

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4 0

3 years ago

How many carbon atoms are there in .500 mol of CO2?

AURORKA [14]

Answer: There are $3.011 \times 10^{23}$ atoms present in 0.500 mol of $CO_{2}$ .

Explanation:

According to the mole concept, there are $6.022 \times 10^{23}$ atoms present in 1 mole of a substance.

In a molecule of $CO_{2}$ there is only one carbon atom present. Therefore, number of carbon atoms present in 0.500 mol of $CO_{2}$ are as follows.

$1 \times 0.500 \times 6.022 \times 10^{23}\\= 3.011 \times 10^{23}$

Thus, we can conclude that there are $3.011 \times 10^{23}$ atoms present in 0.500 mol of $CO_{2}$ .

6 0

2 years ago

In a chemical reaction, 28g of iron reacts with 16g of sulfur to produce _________ of iron sulfide.

kirill115 [55]

Hello!

In a chemical reaction, 28g of iron reacts with 16g of sulfur to produce _________ of iron sulfide.

We have the following data:

m(Fe) - mass of iron = 28 g

m(S) - mass of sufur = 16 g

MM(Fe) - molar mass of iron ≈ 56 g/mol

MM(S) - molar mass of sulfur ≈ 32 g/mol

n(Fe) - number of mol of iron = ?

n(S) - number of mol of sulfur = ?

Solving:

* to n(Fe)

$n_{Fe} = \dfrac{m_{Fe}}{MM_{Fe}}$

$n_{Fe} = \dfrac{28\:\diagup\!\!\!\!\!g}{56\:\diagup\!\!\!\!\!g/mol}$

$\boxed{n_{Fe} = 0.5\:mol}$

* to n(S)

$n_{S} = \dfrac{m_{S}}{MM_{S}}$

$n_{S} = \dfrac{16\:\diagup\!\!\!\!\!g}{32\:\diagup\!\!\!\!\!g/mol}$

$\boxed{n_{S} = 0.5\:mol}$

The stoichiometric reaction will be in the same proportion (1 : 1), let us see:

$Fe + S \Longrightarrow FeS$

1 mol of Fe -------------- 1 mol of FeS

0.5 mol of Fe ------------ 0.5 mol of FeS

Will the reaction of the iron mass with the mass of sulfur produce how many grams of iron sulfide? We will see:

n(FeS) - number of mol of iron sulfide = 0.5 mol

m(FeS) - mass of iron sulfide = ? (in grams)

MM(FeS) - Molar Mass of iron sulfide = 56 + 32 = 88 g/mol

Solving:

$n_{FeS} = \dfrac{m_{FeS}}{MM_{FeS}}$

$m_{FeS} = n_{FeS}*MM_{FeS}$

$m_{FeS} = 0.5\:mol\!\!\!\!\!\!\!\!\!\!\!\dfrac{\hspace{0.6cm}}{~}*88\:\dfrac{g}{mol\!\!\!\!\!\!\!\!\!\!\!\dfrac{\hspace{0.6cm}}{~}}$

$\boxed{\boxed{m_{FeS} = 44\:g}}\:\:\:\:\:\:\bf\blue{\checkmark}\bf\green{\checkmark}\bf\red{\checkmark}$

Answer:

44 grams of iron sulfide

___________________________

$\bf\red{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}$

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3 years ago

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3 years ago

find the percent yield,if 6.0g of Licl are actually produced and the theoretical yield is 35.4g of Licl.​

find the percent yield,if 6.0g of Licl are actually produced and the theoretical yield is 35.4g of Licl.