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4 years ago

How many mole of ZnCl2 will be produced from 55.0 g of Zn, assuming HCL is available in excess

Chemistry

1 answer:

Ilia_Sergeevich [38]4 years ago

8 0

Answer:

0.84 mol

Explanation:

Given data:

Moles of ZnCl₂ produced = ?

Mass of Zn = 55.0 g

Solution:

Chemical equation:

2HCl + Zn → ZnCl₂ + H₂

Number of moles of Zn:

Number of moles = mass / molar mass

Number of moles = 55.0 g/ 65.38 g/mol

Number of moles = 0.84 mol

Now we will compare the moles of Zn with ZnCl₂ from balance chemical equation.

Zn : ZnCl₂

1 : 1

0.84 : 0.84

So from 55 g of Zn 0.84 moles of zinc chloride will be produced.

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Burning a compound of calcium, carbon, and nitrogen in oxygen in a combustion train generates calcium oxide , carbon dioxide , n

mylen [45]

The question is incomplete, here is the complete question:

Burning a compound of calcium, carbon, and nitrogen in oxygen in a combustion train generates calcium oxide (CaO), carbon dioxide $(CO_2)$ , nitrogen dioxide $(NO_2)$ , and no other substances. A small sample gives 2.389 g CaO, 1.876 g $CO_2$ , and 3.921 g $NO_2$ Determine the empirical formula of the compound.

Answer: The empirical formula for the given compound is $CaCN_2$

Explanation:

The chemical equation for the combustion of compound having calcium, carbon and nitrogen follows:

$Ca_xC_yN_z+O_2\rightarrow CaO+CO_2+NO_2$

where, 'x', 'y' and 'z' are the subscripts of calcium, carbon and nitrogen respectively.

We are given:

Mass of CaO = 2.389 g

Mass of $CO_2=1.876g$

Mass of $NO_2=3.921g$

We know that:

Molar mass of calcium oxide = 56 g/mol

Molar mass of carbon dioxide = 44 g/mol

Molar mass of nitrogen dioxide = 46 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 1.876 g of carbon dioxide, $\frac{12}{44}\times 1.876=0.5116g$ of carbon will be contained.

For calculating the mass of nitrogen:

In 46 g of nitrogen dioxide, 14 g of nitrogen is contained.

So, in 3.921 g of nitrogen dioxide, $\frac{14}{46}\times 3.921=1.193g$ of nitrogen will be contained.

For calculating the mass of calcium:

In 56 g of calcium oxide, 40 g of calcium is contained.

So, in 2.389 g of calcium oxide, $\frac{40}{56}\times 2.389=1.706g$ of calcium will be contained.

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Calcium = $\frac{\text{Given mass of Calcium}}{\text{Molar mass of Calcium}}=\frac{1.706g}{40g/mole}=0.0426moles$

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.5116g}{12g/mole}=0.0426moles$

Moles of Nitrogen = $\frac{\text{Given mass of Nitrogen}}{\text{Molar mass of Nitrogen}}=\frac{1.193g}{14g/mole}=0.0852moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0426 moles.

For Calcium = $\frac{0.0426}{0.0426}=1$

For Carbon = $\frac{0.0426}{0.0426}=1$

For Nitrogen = $\frac{0.0852}{0.0426}=2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of Ca : C : N = 1 : 1 : 2

Hence, the empirical formula for the given compound is $CaCN_2$

3 0

3 years ago

Complete the transmutation equation below. Assume that there is only one unknown product. Np93239→Pu94239+?

aliya0001 [1]

I have attached the answer. hopefully, i read the problems correctly. let me know if I did not.

both problems are an example of beta decays. when an atoms' atomic number is increased by one. this is symbolized with -1 e

5 0

3 years ago

Which element shares the most characteristics with magnesium (Mg)?

hjlf

Probably sodium Na because it has a similar atomic mass and number

4 0

3 years ago

What happens when a condition of metal oxide is tested with (i)blue litmus (ii)red litmus

Artemon [7]

it has no effect on the blue litmus.

When it's tested on the red litmus paper it turns blue

7 0

4 years ago

A SCUBA diver with a 4-liter lung capacity is at a depth of 135 feet (4 atm.) She ascends to the surface where the

Nostrana [21]

Answer:

16L

Explanation:

Data obtained from the question include:

V1 (initial volume) = 4L

P1 (initial pressure) = 4atm

P2 (final pressure) = 1atm

V2 (final volume) =?

Using Boyle's law equation P1V1 = P2V2, the new volume can be obtain as follow:

P1V1 = P2V2

4 x 4 = 1 x V2

16 = V2

V2 = 16L

Therefore, her new lungs volume is 16L

5 0

3 years ago

Read 2 more answers