42. consider the reaction: 2 al(oh)₃ + 3 h₂so₄ → al₂(so₄)₃ + 6 h₂o. how many grams of al₂(so₄)₃ are…

Explanation:

Step1: Calculate molar mass

The molar mass of $H_2SO_4$ is $2\times1 + 32+4\times16=98\ g/mol$, and the molar mass of $Al_2(SO_4)_3$ is $2\times27 + 3\times(32 + 4\times16)=342\ g/mol$.

Step2: Find moles of $H_2SO_4$

The number of moles of $H_2SO_4$, $n_{H_2SO_4}=\frac{m}{M}=\frac{152\ g}{98\ g/mol}\approx1.551\ mol$.

Step3: Determine mole - ratio

From the balanced chemical equation $2Al(OH)_3 + 3H_2SO_4\rightarrow Al_2(SO_4)_3+6H_2O$, the mole - ratio of $H_2SO_4$ to $Al_2(SO_4)_3$ is $3:1$. So the number of moles of $Al_2(SO_4)3$, $n{Al_2(SO_4)3}=\frac{1}{3}n{H_2SO_4}=\frac{1}{3}\times1.551\ mol\approx0.517\ mol$.

Step4: Calculate mass of $Al_2(SO_4)_3$

The mass of $Al_2(SO_4)_3$, $m = n\times M=0.517\ mol\times342\ g/mol\approx177\ g$.

Answer:

C. 177 g of $Al_2(SO_4)_3$