Question

Correctly order the steps necessary to calculate the mass of product formed in a chemical reaction given the masses of all reactants

Answer 1

1) write down balanced chemical reaction.
2) calculate the number of moles of a reagents.
3) determining molar ratios - identify the number of moles of each reactant needed to form a certain number of moles of each product.
4) from number of moles of each product calculate mass of products using molar mass.

Answer 2

Answer : The order of steps necessary to calculate the mass of product formed in a chemical reaction given the masses of all reactants :

(1) Determine the number of moles of each reactant by dividing its mass by its molar mass.

(2) Identify the limiting reactant as the reactant that produces the least amount of product.

(3) Determine the amount of product that could be formed from the limiting reactant using appropriate mole ratio.

(4) Determine the mass, multiply the number of moles of product formed (from the limiting reactant) by the molar mass of the product.

For example :

If we are given that 10 grams of magnesium react with 6 grams of oxygen to give magnesium oxide then calculate the mass of product formed in a chemical reaction.

Solution :

Mass of Mg = 10 g

Mass of $O_2$ = 6 g

Molar mass of Mg = 24 g/mole

Molar mass of $O_2$ = 32 g/mole

Molar mass of MgO = 40 g/mole

First we have to calculate the moles of Mg and $O_2$.

$\text{ Moles of }Mg=\frac{\text{ Mass of }Mg}{\text{ Molar mass of }Mg}=\frac{10g}{24g/mole}=0.4167moles$

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{6g}{32g/mole}=0.1875moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$2Mg(s)+O_2(g)\rightarrow 2MgO(s)$

From the balanced reaction we conclude that

As, 1 mole of $O_2$ react with 2 mole of $Mg$

So, 0.1875 moles of $O_2$ react with $0.1875\times 2=0.375$ moles of $Mg$

From this we conclude that, $Mg$ is an excess reagent because the given moles are greater than the required moles and $O_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $MgO$

From the reaction, we conclude that

As, 1 mole of $O_2$ react to give 2 mole of $MgO$

So, 0.1875 moles of $O_2$ react to give $0.1875\times 2=0.375$ moles of $MgO$

Now we have to calculate the mass of $MgO$

$\text{ Mass of }MgO=\text{ Moles of }MgO\times \text{ Molar mass of }MgO$

$\text{ Mass of }MgO=(0.375moles)\times (40g/mole)=15g$

Hence, the mass of product is 15 grams.