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
