Question

question 8 10 pts aluminum hydroxide is one of the ingredients used in over the counter antacids - it lowers the amount of acid (hcl) in a stomach by neutralizing it and producing aluminum chloride and water. how many grams of hcl can be neutralized by 0.200 g of aluminum hydroxide? the equation for the reaction is al(oh)₃ + 3 hcl → alcl₃ + 3 h₂o 0.0769 g 0.0935 g 0.280 g 1.28 g question 9 10 pts when 4.03 g of hydrogen gas and 32.00 g of oxygen gas react, it is expected that 36.03 g of water will be produced. a student has conducted this experiment and obtained only 29.50 g of water. what was the percent yield of the experiment? 81.88% 36.03% 29.50% 122.1%

Explanation:

Step1: Calculate moles of aluminum hydroxide

The molar mass of $Al(OH)_3$ is $27+(16 + 1)\times3=78$ g/mol. Moles of $Al(OH)_3=\frac{0.200\ g}{78\ g/mol}=\frac{0.200}{78}$ mol.

Step2: Determine mole - ratio from the balanced equation

From $Al(OH)_3 + 3HCl
ightarrow AlCl_3+3H_2O$, the mole - ratio of $Al(OH)_3$ to $HCl$ is 1:3. So moles of $HCl = 3\times\frac{0.200}{78}$ mol.

Step3: Calculate mass of HCl

The molar mass of $HCl$ is $1 + 35.5 = 36.5$ g/mol. Mass of $HCl=3\times\frac{0.200}{78}\text{ mol}\times36.5\text{ g/mol}\approx0.280$ g.

Step4: Calculate percent - yield for the second question

Percent yield = $\frac{\text{Actual yield}}{\text{Theoretical yield}}\times100\%$. Given actual yield of water = 29.50 g and theoretical yield = 36.03 g. Percent yield=$\frac{29.50\ g}{36.03\ g}\times100\%\approx81.88\%$.

Answer:

Question 8: 0.280 g
Question 9: 81.88%