Stoichiometry problem-
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Stoichiometry problems-
Question 1-
From the reaction: B2H6 + O2
HBO2 + H2O
a. What mass of O2 will be needed to burn 36.1 g of B2H6?
b. How many moles of water are produced from 19.2 g of B2H6?
Ans-
Part a) Balanced reaction is,
B2H6 + 3O2 2HBO2 + 2H2O
molar mass of B2H6= 27.67 gm/mole
mass of B2H6= 36.1 gm
moles of B2H6 = mass / molar mass= 36.1 / 27.67= 1.305 mole
1 mole B2H6 reacts with = 3 mole O2
1.305 mole B2H6 reacts with = 3x 1.305 mole O2
moles of O2= 3.915
molar mass of O2 = 32 gm/mole
mass of O2 = moles x molar mass = 3.915 x 32 = 125.28 gm
Mass of O2 will be needed to burn 36.1 g of B2H6 = 125.28 gm
Part b)
molar mass of B2H6= 27.67 gm/mole
mass of B2H6= 19.2 gm
moles of B2H6 = mass / molar mass= 19.2 / 27.67= 0.694 mole
1 mole B2H6 on reaction gives = 2 mole H2O
0.694 mole B2H6 on reaction gives =2 x 0.694 mole H2O
moles of H2O = 1.388 mole
1.388 moles of water are produced from 19.2 g of B2H6.
Stoichiometry problem-
Question 2- Calculate the mass (in kg) of water produced from the combustion of 1.0 gallon (3.8 L) of gasoline (C8H18). The density of gasoline is 0.79 g/mL.
Answer –
2C8H18 + 25O2 —> 16CO2 + 18H2O
2(12×8 + 18×1) 18( 1×2+16)
= 228 gm = 324 gm
Volume of gasoline = 1 gallon = 3.8 L = 3800 ml
density of gasoline= 0.79 gm/ml
mass of gasoline = V xd = 3800 x 0.79 =3002 gm
228 gm gasoline on combustion gives = 324 gm H2O
3002 gm gasoline on combustion gives = 324 x 3002 / 228 gm H2O = 4266 gm
mass of water = 4266 gm
Mass of water produced = 4.266 Kg
Stoichiometry problem-
Question 3-
One mole of aspartame (C14H18N2O5) reacts with two moles of water to produce one mole of aspartic acid (C4H7NO4), one mole of methanol (CH3OH) & one mole of phenylalanine.
a. What is the molecular formula of phenylalanine?
b. What mass of phenylalanine is produced from 378 g of aspartame?
Answer-
Part a)
C14H18N2O5 + 2 H2O —> C4H7NO4 + CH3OH + phenylalanine
From the above- balanced equation formula of phenylalanine is C9H11NO2.
C14H18N2O5 + 2 H2O —> C4H7NO4 + CH3OH +C9H11NO2
Part b)
molar mass of aspartame= 294.3 gm /mole
mass of aspartame = 378 gm
moles = 378 / 294.3 = 1.28
1 mole aspartame gives = 1 mole phenylalanine
1.28 mole aspartame gives = 1 x 1.28 = 1.28 mole phenylalanine
molar mass of phenylalanine = 165.19 gm/mole
mass of of phenylalanine = moles x molar mass = 1.28 x 165.19 = 211.44 gm
211.44 gm mass of phenylalanine is produced from 378 g of aspartame.