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dimulka [17.4K]
3 years ago
8

How is stoichiometry used to calculate the amount of product from amount of reactant?

Chemistry
2 answers:
Nesterboy [21]3 years ago
7 0

Stoichiometry relates moles of rectant to moles of product

IceJOKER [234]3 years ago
6 0
<span>Stoichiometry deals with the quantitative measurement of reactants and products in a chemical reaction. Let suppose you are given with following reaction;

                                            A  +  2 B   </span>→    3 C

According to this reaction 1 mole of A reacts with 2 moles of B to produce 3 moles of C. Now using the concept of mole one can easily measure the amount of reactants reacted and the amount of product formed, as...

                    1 Mole Exactly equals 6.022 × 10²³ particles

                    1 Mole of Gas (at STP) exactly occupies 22.4 L Volume

                    1 Mole of any compound exactly equals the molar mass in                                 grams

Therefore, <span>Stoichiometry is very helpful in quantitative analysis.</span>
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A 1.555-g sample of baking soda decomposes with heat to produce 0.991 g Na2CO3. Refer to Example Exercise 14.l and show the calc
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Answer:

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Explanation:

The equation for the thermal decomposition of baking soda is shown;

2NaHCO3 → Na2CO3 + H2O + CO2

Number of moles of baking soda= mass/molar mass= 1.555g/84.007 g/mol = 0.0185 moles

From the reaction equation;

2 moles of baking soda yields 1 mole of sodium carbonate

0.0185 moles of baking soda will yield = 0.0185 moles ×1 /2 = 9.25 ×10^-3 moles of sodium carbonate.

Therefore, mass of sodium carbonate= 9.25 ×10^-3 moles × 106gmol-1= 0.9805 g of sodium carbonate. This is the theoretical yield of sodium carbonate.

%yield = actual yield/theoretical yield ×100

% yield = 0.991/0.9805 ×100

%yield = 101%

Since ;

2NaHCO3 → Na2CO3 + H2O + CO2

And H2O + CO2 ---> H2CO3

Hence I can write, 2NaHCO3 → Na2CO3 + H2CO3

Molar mass of H2CO3= 62.03 gmol-1

Molar mass of baking soda= 84 gmol-1

Therefore, mass of baking soda=

0.325/62.03 × 2 × 84 = 0.88 g of NaHCO3

% of NaHCO3= 0.88/1.473 × 100 = 59.7%

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