Answer:
312.46 grams of aluminum are required to produce 17.5 grams of hydrogen.
Step-by-step explanation:
To answer this question, we need to know the chemical equation for the reaction between aluminum and hydrogen.
2 Al + 6 HCl → 2 AlCl3 + 3 H2
This equation shows that 2 moles of aluminum react with 3 moles of hydrogen to produce 3 moles of H2 gas.
From the equation, we can see that the molar ratio of aluminum to hydrogen is 2:3. That means for every 2 moles of aluminum, we get 3 moles of hydrogen.
To find out how many grams of aluminum are required to produce 17.5 grams of hydrogen, we need to use the molar mass of hydrogen and the molar ratio of aluminum to hydrogen.
The molar mass of hydrogen is 1.008 g/mol.
First, we need to calculate the number of moles of hydrogen in 17.5 grams:
moles of H2 = mass / molar mass
moles of H2 = 17.5 g / 1.008 g/mol
moles of H2 = 17.36 mol
Next, we can use the molar ratio to find the number of moles of aluminum required:
moles of Al = (2/3) * moles of H2
moles of Al = (2/3) * 17.36 mol
moles of Al = 11.57 mol
Finally, we can calculate the mass of aluminum required:
mass of Al = moles of Al * molar mass of Al
mass of Al = 11.57 mol * 26.982 g/mol
mass of Al = 312.46 g
Therefore, 312.46 grams of aluminum are required to produce 17.5 grams of hydrogen.