Question

3.) h₂o + so₃ → h₂so₄ 15g blankg → 60 g
7.) na₂ + h₂so₄ → na₂so₄ + h₂ 29 g blankg → 46 g 4g
9.) cu + 2h₂so₄ → cuso₄ + 2h₂o + so₂ blankg 15 g → 35g 12g 3g

Explanation:

Step1: Apply the law of conservation of mass

The law of conservation of mass states that the total mass of reactants equals the total mass of products in a chemical reaction.

For the reaction $H_2O + SO_3

ightarrow H_2SO_4$:
Let the mass of $SO_3$ be $x$.
We know that mass of $H_2O = 15g$ and mass of $H_2SO_4=60g$.
According to the law of conservation of mass, $15 + x=60$.

Step2: Solve for $x$

$x = 60 - 15$.
$x = 45g$.

For the reaction $Na_2 + H_2SO_4

ightarrow Na_2SO_4 + H_2$:
Let the mass of $H_2SO_4$ be $y$.
We know that mass of $Na_2 = 29g$, mass of $Na_2SO_4 = 46g$ and mass of $H_2=4g$.
According to the law of conservation of mass, $29 + y=46 + 4$.

Step3: Solve for $y$

$29 + y=50$.
$y=50 - 29$.
$y = 21g$.

For the reaction $Cu + 2H_2SO_4

ightarrow CuSO_4+2H_2O + SO_2$:
Let the mass of $Cu$ be $z$.
We know that mass of $H_2SO_4 = 15g$, mass of $CuSO_4 = 35g$, mass of $H_2O = 12g$ and mass of $SO_2=3g$.
According to the law of conservation of mass, $z + 15=35 + 12+3$.

Step4: Solve for $z$

$z+15 = 50$.
$z=50 - 15$.
$z = 35g$.

Answer:

3.) $45$
7.) $21$
9.) $35$