In: Chemistry
For each of the following reactions, calculate the grams of indicated product when 16.0 g of the first reactant and 10.5 g of the second reactant is used:
A). 4Li(s)+O2(g)→2Li2O(s) (Li2O)
B). Fe2O3(s)+3H2(g)→2Fe(s)+3H2O(l) (Fe)
C). Al2S3(s)+6H2O(l)→2Al(OH)3(aq)+3H2S(g) (H2S)
A)
Molar mass of Li = 6.968 g/mol
mass(Li)= 16.0 g
use:
number of mol of Li,
n = mass of Li/molar mass of Li
=(16 g)/(6.968 g/mol)
= 2.296 mol
Molar mass of O2 = 32 g/mol
mass(O2)= 10.5 g
use:
number of mol of O2,
n = mass of O2/molar mass of O2
=(10.5 g)/(32 g/mol)
= 0.3281 mol
Balanced chemical equation is:
4 Li + O2 ---> 2 Li2O
4 mol of Li reacts with 1 mol of O2
for 2.296 mol of Li, 0.5741 mol of O2 is required
But we have 0.3281 mol of O2
so, O2 is limiting reagent
we will use O2 in further calculation
Molar mass of Li2O,
MM = 2*MM(Li) + 1*MM(O)
= 2*6.968 + 1*16.0
= 29.936 g/mol
According to balanced equation
mol of Li2O formed = (2/1)* moles of O2
= (2/1)*0.3281
= 0.6562 mol
use:
mass of Li2O = number of mol * molar mass
= 0.6562*29.94
= 19.65 g
Answer: 19.7 g
B)
Molar mass of Fe2O3,
MM = 2*MM(Fe) + 3*MM(O)
= 2*55.85 + 3*16.0
= 159.7 g/mol
mass(Fe2O3)= 16.0 g
use:
number of mol of Fe2O3,
n = mass of Fe2O3/molar mass of Fe2O3
=(16 g)/(1.597*10^2 g/mol)
= 0.1002 mol
Molar mass of H2 = 2.016 g/mol
mass(H2)= 10.5 g
use:
number of mol of H2,
n = mass of H2/molar mass of H2
=(10.5 g)/(2.016 g/mol)
= 5.208 mol
Balanced chemical equation is:
Fe2O3 + 3 H2 ---> 2 Fe + 3 H2O
1 mol of Fe2O3 reacts with 3 mol of H2
for 0.1002 mol of Fe2O3, 0.3006 mol of H2 is required
But we have 5.208 mol of H2
so, Fe2O3 is limiting reagent
we will use Fe2O3 in further calculation
Molar mass of Fe = 55.85 g/mol
According to balanced equation
mol of Fe formed = (2/1)* moles of Fe2O3
= (2/1)*0.1002
= 0.2004 mol
use:
mass of Fe = number of mol * molar mass
= 0.2004*55.85
= 11.19 g
Answer: 11.2 g
C)
Molar mass of Al2S3,
MM = 2*MM(Al) + 3*MM(S)
= 2*26.98 + 3*32.07
= 150.17 g/mol
mass(Al2S3)= 16.0 g
use:
number of mol of Al2S3,
n = mass of Al2S3/molar mass of Al2S3
=(16 g)/(1.502*10^2 g/mol)
= 0.1065 mol
Molar mass of H2O,
MM = 2*MM(H) + 1*MM(O)
= 2*1.008 + 1*16.0
= 18.016 g/mol
mass(H2O)= 10.5 g
use:
number of mol of H2O,
n = mass of H2O/molar mass of H2O
=(10.5 g)/(18.02 g/mol)
= 0.5828 mol
Balanced chemical equation is:
Al2S3 + 6 H2O ---> 3 H2S + 2 Al(OH)3
1 mol of Al2S3 reacts with 6 mol of H2O
for 0.1065 mol of Al2S3, 0.6393 mol of H2O is required
But we have 0.5828 mol of H2O
so, H2O is limiting reagent
we will use H2O in further calculation
Molar mass of H2S,
MM = 2*MM(H) + 1*MM(S)
= 2*1.008 + 1*32.07
= 34.086 g/mol
According to balanced equation
mol of H2S formed = (3/6)* moles of H2O
= (3/6)*0.5828
= 0.2914 mol
use:
mass of H2S = number of mol * molar mass
= 0.2914*34.09
= 9.933 g
Answer: 9.93 g