Stoichiometry problem-

Stoichiometry problems

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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.