Stoichiometry

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## Stoichiometry Numericals-

## Question 1-

## How many moles & atoms are in the following-

## (i) 14.01 gm N (ii) 80.16 gm Ca (iii) 76.41 Vanadium

## Solution-

** (i) 14.01 gm N**

mass of N = 14.01 gm,molar mass of N = 14.01 gm/mole

moles = mass / molar mass

= 14.01 / 14.01

**mole of N = 1 mole**

1 mole of N has = 6.023x 1023 atoms

**no. of N atoms = 6.023x 1023**

** (ii) 80.16 gm Ca**

mass of Ca = 80.16 gm,molar mass of Ca = 40.08 gm/mole

moles = mass / molar mass

= 80.16/ 40.08

**mole of Ca = 2 mole**

1 mole of Ca has = 6.023x 1023 atoms (avogadro’s no.)

2 mole of Ca has = 2 x 6.023x 1023 atoms

**no. of Ca atoms = 12.046 x 1023**

**(iii) 76.41 gm Vanadium
**

mass of V = 76.41 gm,molar mass of V = 50.94 gm/mole

moles = mass / molar mass

=76.41 / 50.94

**mole of V = 1.5 mole**

1 mole of V has = 6.023x 1023 atoms (avogadro’s no.)

1.5 mole of V has = 1.5 x 6.023x 1023 atoms

**no. of V atoms = 9.0345 x 1023**

## Question 2- How many Mg^{+2 } ions are found in 1.00 mole of MgO ?

### Ans-

1.0 mole MgO = 1.0 mole Mg^{+2 }

Therefore,

moles of Mg^{+2 }= 1.0

we know,

1.0 mole Mg^{+2}^{ }= Avogadro’s number

**No. of Mg ^{+2}^{ }in 1.0 mole MgO = 6.022 x 10 ^{23 }Mg^{+2}^{ } ions**

## Stoichiometry Numericals-

### Question 3- How many grams of Al can be created by decomposition of 9.8 gm of Al2O3 ?

### Solution –

2Al2O3 —> 4Al + 3O2

2 mole 4 mole 3 mole

mass of Al2O3 = 9.8 gm

molar mass of Al2O3= 102 gm/mole

moles of Al2O3 = mass / molar mass

= 9.8 / 102

moles of Al2O3 = 0.0961

2 moles Al2O3 gives = 4 mole Al

0.0961 moles Al2O3 gives = 4 x 0.0961 / 2 = 0.1922 mole Al

moles of Al = 0.1922

molar mass of Al = 27 gm /mole

mass of Al = moles x molar mass = 0.1922 x 27 = 5.1894

**Mass of Al = 5.19 gm**

### Question 4-

### 0.5 L of 0.20 M HCl solution is mixed with 0.5L of 0.40M AgNO3 solution & reaction occurs.If the reaction goes to completion ,what mass of AgCl produces ?

### Ans-

HCl + AgNO3 —> AgCl + HNO3

Molarity of HCl = 0.20M

Volume of HCl = 0.5L

moles of HCl = Molarity x volume(L)

= 0.20 x 0.5

**moles of HCl = 0.1**

Molarity of AgNO3= 0.40M

Volume of AgNO3 = 0.5L

moles of AgNO3 = Molarity x volume(L)

= 0.40 x 0.5

moles of AgNO3 = 0.2

1mole HCl reacts with= 1 mole AgNO3

0.1mole HCl reacts with= 0.1 mole AgNO3

But 0.2 mole AgNO3 are present ,so HCl is limiting reagent.

**0.1 mole HCl gives 0.1 mole AgCl.**

molar mass of AgCl = 143.32 gm/mole

mass of AgCl = mole x molar mass = 0.1 x 143.32 = 14.332 gm

**Mass of AgCl = 14.332 gm**