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3 years ago

The gram formula mass of NH4Cl is(1) 22.4 g/mole (3) 53.5 g/mole(2) 28.0 g/mole (4) 95.5 g/mole

2 answers:

7 0

The answer is (3) 53.5 g/mol. The gram formula mass means that the mass of one mol compound. 1N=14, 4H=4, 1Cl=35.5. So the gram formula mass of NH4Cl=14+4+35.5=53.5 g/mol.

qwelly [4]3 years ago

6 0

Answer: Option (3) is the correct answer.

Explanation:

Gram formula mass or molar mass is the sum of mass of all the individual atoms present in the compound or formula.

For example, gram formula mass of $NH_{4}Cl$ is calculated as follows.

$NH_{4}Cl = Mass of N + 4 \times mass of H + mass of Cl$

= $(14.00 + 4 \times 1.00 + 35.45)$ g/mol

= (14.00 + 4.00 + 35.45) g/mol

= 53.45 g/mol

= 53.5 g/mol (approx)

Thus, we can conclude that the gram formula mass of $NH_{4}Cl$ is 53.5 g/mol (approx)

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A certain gas at 2oC and 1.00 atm pressure fills a 4.0 container. What volume will the gas occupy at 100oC and 780 torr pressure

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<h3>Further explanation</h3>

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3 years ago

onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency

mestny [16]

Answer : The excess reactant is, $H_2O$

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of $CaCN_2$ = 105.0 g

Mass of $H_2O$ = 78.0 g

Molar mass of $CaCN_2$ = 80.11 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $CaCO_3$ = 100.09 g/mole

First we have to calculate the moles of $CaCN_2$ and $H_2O$ .

$\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles$

$\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $CaCN_2$ react with 3 mole of $H_2O$

So, 1.31 moles of $CaCN_2$ react with $1.31\times 3=3.93$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $CaCN_2$ is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

$\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)$

$\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g$

Thus, the leftover amount of excess reagent is, 7.2 grams.

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