2 aluminum reacts with sulfuric acid to produ...
2 aluminum reacts with sulfuric acid to produce aluminum sulfate and hydrogen gas. how many grams (rounded to two decimal places) of aluminum sulfate would be formed if 250 g of sulfuric acid completely reacted with aluminum?
Answer
# Explanation:
## Step1: Write the balanced chemical equation
$2Al + 3H_2SO_4=Al_2(SO_4)_3 + 3H_2$
## Step2: Calculate the molar mass of $H_2SO_4$ and $Al_2(SO_4)_3$
The molar mass of $H_2SO_4$: $M_{H_2SO_4}=2\times1 + 32+4\times16=98\ g/mol$.
The molar mass of $Al_2(SO_4)_3$: $M_{Al_2(SO_4)_3}=2\times27 + 3\times(32 + 4\times16)=342\ g/mol$.
## Step3: Calculate the number of moles of $H_2SO_4$
$n_{H_2SO_4}=\frac{m_{H_2SO_4}}{M_{H_2SO_4}}=\frac{250\ g}{98\ g/mol}\approx2.551\ mol$.
## Step4: Determine the mole - ratio between $H_2SO_4$ and $Al_2(SO_4)_3$
From the balanced equation, the mole - ratio of $H_2SO_4$ to $Al_2(SO_4)_3$ is $3:1$. So the number of moles of $Al_2(SO_4)_3$ formed, $n_{Al_2(SO_4)_3}=\frac{1}{3}n_{H_2SO_4}$.
$n_{Al_2(SO_4)_3}=\frac{1}{3}\times2.551\ mol\approx0.8503\ mol$.
## Step5: Calculate the mass of $Al_2(SO_4)_3$
$m_{Al_2(SO_4)_3}=n_{Al_2(SO_4)_3}\times M_{Al_2(SO_4)_3}=0.8503\ mol\times342\ g/mol\approx290.80\ g$.
# Answer:
$290.80\ g$