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BlackZzzverrR [31]

3 years ago

5

if you react 100g of ammonium chloride with excess calcium oxide, what Is the theoretical yeild (in grams) of ammonia? when you

complete the reaction, your actual yeild of ammonia is 8.12g. what is your percent yeild for the production of ammonia using this process

1 answer:

Maurinko [17]3 years ago

4 0

Answer:

31.78 grams

25.55%

Explanation:

The balanced reaction for ammonium chloride with calcium oxide will be:

2NH4Cl + Ca(OH)2 ---> CaCl2 + 2NH3 + 2H2O

The molecular weight for ammonium chloride(NH4Cl ) is 53.49g/mol, while the molecular weight for ammonium(NH3) is 17g/mol. The number of theoretical yield of ammonia from 100g of ammonium chloride will be:

100g / (53.49g/mol) * 2/2 * 17g/mol= 31.78 grams

If the actual yield is 8.12g, the percent yield will be: 8.12g/31.78g * 100% =25.55%

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The formula for binary ionic compound formed between cesium and fluorine

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Answer:

Cesium fluoride(CsF)

Explanation:

A binary compound is a compound that is composed of 2 distinct element. An ionic compound is composed of ions, usually one is a metal why the other is a non metal. One element gives out electron to form cation and the other receives electron to form anion in a binary compound.

Cesium is a group 1 element and it has one valence electron and it can easily donate this 1 electron to form a bond with other element. Group 1 element are generally very reactive. Cesium is a metal

Fluorine is in group 7 of the periodic table and is a non metal .Fluorine have 7 valency electron and requires 1 electron to form a stable octet.

When cesium and fluorine bond to form a binary compound cesium donate 1 electron and fluorine receives the 1 electron for both element to form a stable octet. The formula for the binary ionic compound of cesium and fluorine can therefore be expressed as Cesium fluoride(CsF)

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A compound is composed of C, H and O. A 1.621 g sample of this compound was combusted, producing 1.902 g of water and 3.095 g of

vlada-n [284]

Answer: The molecular of the compound is, $C_2H_3O$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=3.095g$

Mass of $H_2O=1.902g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

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

So, in $3.095g$ of carbon dioxide, $\frac{12}{44}\times 3.095=0.844g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in $1.902g$ of water, $\frac{2}{18}\times 1.092=0.121g$ of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = $(1.621)-[(0.844)+(0.121)]=0.656g$

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.844g}{12g/mole}=0.0703moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.121g}{1g/mole}=0.121moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.656g}{16g/mole}=0.041moles$

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 $0.041$ moles.

For Carbon = $\frac{0.0703}{0.041}=1.71\approx 2$

For Hydrogen = $\frac{0.121}{0.041}=2.95\approx 3$

For Oxygen = $\frac{0.041}{0.041}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 3 : 1

Hence, the empirical formula for the given compound is $C_2H_3O_1=C_2H_3O$

The empirical formula weight = 2(12) + 3(1) + 1(16) = 43 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{46.06}{43}=1$

Molecular formula = $(C_2H_3O_1)_n=(C_2H_3O_1)_1=C_2H_3O$

Therefore, the molecular of the compound is, $C_2H_3O$

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2 years ago