CBSE Class 11-science Answered

Hello sir, please clarify my doubt. can you explain me in detail about stoichiometry and limiting reagents with examples for each. Thank you.

Asked by seeni2005 | 28 Jul, 2020, 11:03: AM

Expert Answer

Dear Student,

Please follow the given link to understand the concept and stoichiometry-

https://images.topperlearning.com/topper/revisionnotes/3709_Topper_21_101_3_2_27_34_Mass_and_Stoichiometry_up201605111103_1462944788_0608.pdf?v=0.0.1

Now, Let's have a look on examples to solve the problems-

Question-1:

What is the amount of magnesium required to form 160 gram of magnesium oxide by burning magnesium in air?

Solution:

Mass of magnesium oxide = 160 gm

Reaction equation is;

space space space space space space 2 Mg space space plus space space straight O subscript 2 space space rightwards arrow space space 2 MgO space space space 2 space mole space space space space space 1 space mole space space space space 2 space mole Moles space of space MgO space equals fraction numerator Mass space of space MgO over denominator Molar space mass end fraction space space space space space space space space space space space space space space space space space space space space space space equals 160 over 40 Moles space of space MgO space equals space 4 space moles 2 space moles space og space MgO space requires space 2 space mole space of space Mg 4 space moles space og space MgO space will space require space space fraction numerator 4 cross times 2 over denominator 2 end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 4 space moles space of space Mg

4 moles of magnesium is required to form 160 gm of magnesium oxide.

Question -2:

How many moles of Al2O3 will be formed when a mixture of 5.4g Al and 3.2g O2 is heated?

Kindly provide the answer.

Solution-

$Given colon space Mass space of space Al space equals space 5.4 space straight g Mass space of space straight O subscript 2 space equals space 3.2 space straight g. Balanced space equation colon 4 Al space plus space 3 space straight O subscript 2 space space rightwards arrow space 2 Al subscript 2 straight O subscript 3 Now comma Moles space of space Al space equals space fraction numerator Mass space over denominator Molar space mass end fraction space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 5.4 over denominator 26.98 end fraction Moles space of space Al space space equals space 0.2 space mol Moles space of space straight O subscript 2 equals space fraction numerator 3.2 over denominator 32 end fraction Moles space of space straight O subscript 2 space equals space 0.1 space mol Now comma space space to space find space the space limliting space agent space divide space moles space of space reactants space by space stoichiometric space coefficients For space straight O subscript 2 space equals space fraction numerator 0.1 over denominator 3 end fraction space space space space space space space space space space space space space space space equals space 0.033 For space Al space equals fraction numerator 0.2 over denominator 2 end fraction space space space space space space space space space space space equals space space 0.05 As space value space of space straight O subscript 2 space is space less space than space Al comma space straight O subscript 2 space is space the space limiting space agent. From space reaction colon The space mole space of space straight O subscript 2 space colon space Moles space of space space Al subscript 2 straight O subscript 3 space space space space space space space space space space space space space space space space space space space space space space 3 space colon space 2 therefore space space space space space space space space space space space space space space space space 0.1 space colon space straight x space space space space space space space space space space space space space space space space space straight x space equals space space 0.066 space mol space of space space Al subscript 2 straight O subscript 3$