Question

When a solution of barium nitrate and a solution of copper (II) sulfate are mixed, a chemical reaction produces solid barium sulfate, which sinks to the bottom, and a solution of copper (II) nitrate. Suppose some barium nitrate is dissolved in 120.00 g of water and 8.15 g of copper (II) sulfate is dissolved in 75.00 g of water. The solutions are poured together, and a white solid forms. After the solid is filtered off, it is found to have a mass of 10.76 g. The mass of the solution that passed through the filter is 204.44 g. What mass of barium nitrate was used in the reaction?

Answer 1

The total mass of the products is 10.76 g + 204.44 g = 215.20 g.
The masses of all the reactants but one are known so,

215.20 g - 120.00 g - 8.15 g - 75.00 g = 12.05 g

12.05 g is the mass of the unweighed barium nitrate. 

Answer 2

Mass of barium nitrate used in the beginning of the reaction was $\boxed{12.05{\text{ g}}}$.

Further explanation:

The chemical reaction that contains equal number of atoms of the different elements in the reactant as well as in the product side is known as balanced chemical reaction. The chemical equation is required to be balanced to follow the Law of the conservation of massThe chemical reaction that contains equal number of atoms of the different elements in the reactant as well as in the product side is known as balanced chemical reaction. The chemical equation is required to be balanced to follow the Law of the conservation of mass.

Every chemical reaction occurs in accordance with the law of conservation of mass that states that total mass of the reactant combining must be equal to the products formed. Hence every balanced chemical equation must obey the law of conservation of mass and energy.

The balanced chemical equation for reaction between barium nitrate and copper sulfate is given as follows:

${\text{BaN}}{{\text{O}}_3}\left( l \right) + {\text{CuS}}{{\text{O}}_4}\left( l \right) \to {\text{BaS}}{{\text{O}}_4}\left( s \right) + {\text{CuN}}{{\text{O}}_3}\left( l \right)$  

Since the mass of the solution that passed through the filter was $215.20{\text{ g}}$, and the mass of solid ${\text{BaS}}{{\text{O}}_4}$ precipitate separated was  ${\text{10}}{\text{.76 g}}$. Hence total mass of products is the sum of these two reactants.

$\begin{aligned}{\left( {{\text{Total}}\;{\text{Mass}}} \right)_{{\text{products}}}} &= {\text{10}}{\text{.76 g}} + 215.20{\text{ g}}\\&= 215.20{\text{ g}}\\\end{aligned}$

Assume the mass of unknown barium nitrate is $x$ . Mass of reactants involves mass of barium nitrate, water used for dissolving barium nitrate, mass of copper sulfate and water used for dissolving copper sulfate. The total mass of reactants is calculated as follows:

 $\begin{aligned}{\left( {{\text{Total Mass}}} \right)_{{\text{reactants}}}}&= x + {\text{8}}{\text{.15 g}} + {\text{120}}{\text{.0 g}} + {\text{75}}{\text{.00 g}}\\&= x + 203.15\;{\text{g}}\\\end{aligned}$

The law of conservation of mass holds true for this equation as well as represented below:

${\left( {{\text{Total Mass}}} \right)_{{\text{reactants}}}} = {\left( {{\text{Total}}\;{\text{Mass}}} \right)_{{\text{products}}}}$                                     …… (1)

Substitute $215.20{\text{ g}}$ for ${\left( {{\text{Total}}\;{\text{Mass}}} \right)_{{\text{products}}}}$ and $x + 203.15\;{\text{g}}$ for ${\left( {{\text{Total Mass}}} \right)_{{\text{reactants}}}}$ in equation

(1).

 $x + {\text{120}}{\text{.0 g}} + {\text{75}}{\text{.00 g}} + {\text{8}}{\text{.15 g}} = 215.20{\text{ g}}$

Simplify to obtain the value of mass of barium nitrate $\left( x \right)$.

 $\begin{aligned}x&=215.20{\text{ g}}- {\text{120}}{\text{.0 g}} - {\text{75}}{\text{.00 g}} - {\text{8}}{\text{.15 g}}\\&= 215.20{\text{ g}} - {\text{203}}{\text{.15 g}}\\&= 12.05{\text{ g}}\\\end{aligned}$

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Answer details:  

Grade: High School

Subject: Chemistry

Chapter: Chemical equations and reactions.

Keywords: Barium nitrate, precipitation, precipitate, law of conservation of mass, mass, products, reactants and 12.05 g.