0.16 moles of Aluminum hydroxide are needed to react with 25 grams of sulfuric acid.
Explanation:
In order to find the number of moles we first need to write down the balanced equation as,
[tex]3 H_{2} S O_{4}+2 A l(O H)_{3} \stackrel{\text { yiek }}{\longrightarrow} \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}+6 \mathrm{H}_{2} \mathrm{O}[/tex]
Now as mentioned above,For 3 moles of sulfuric acid , we need 2 moles of Aluminum hydroxide to balance the equation,
Thus we can balance it as,
[tex]\begin{aligned}25 \mathrm{g} \text { of } \mathrm{H}_{2} \mathrm{SO}_{4} & \times \frac{1 \mathrm{mol} \text { of } \mathrm{H}_{2} \mathrm{SO}_{4}}{98.079 \frac{\mathrm{g}}{\mathrm{mol}}} \times \frac{2 \mathrm{mol} \mathrm{Al}(\mathrm{OH})_{3}}{3 \mathrm{mol} \mathrm{H}_{2} \mathrm{SO}_{4}} \\\\&=0.16 \mathrm{mol} \mathrm{Al}(\mathrm{OH})_{3}\end{aligned}[/tex]
Thus it is now clearly known that 0.16 moles of Aluminum hydroxide are needed to react with 25 grams of sulfuric acid.
