When 30.0g of aluminum reacts with HCl, how many grams of aluminum chloride are formed?
Al + HCl --> AlCl3 + H2
To determine the number of grams of aluminum chloride formed when 30.0g of aluminum reacts with HCl, you need to use the concept of a balanced chemical equation and the law of conservation of mass.
1. Start by writing a balanced chemical equation for the reaction between aluminum and HCl:
Al + 3HCl → AlCl3 + H2
2. Calculate the molar mass of aluminum (Al) and aluminum chloride (AlCl3):
Molar mass of Al = 26.98 g/mol
Molar mass of AlCl3 = (27.0 g/mol) + (3 * 35.45 g/mol) = 133.34 g/mol
3. Convert the given mass of aluminum (30.0g) to moles by dividing it by the molar mass of aluminum:
Moles of Al = 30.0 g / 26.98 g/mol = 1.111 mol (rounded to three decimal places)
4. Use the balanced chemical equation to determine the molar ratio between aluminum and aluminum chloride:
From the equation, we see that 1 mole of Al reacts with 1 mole of AlCl3.
5. Calculate the moles of aluminum chloride formed using the molar ratio determined in step 4:
Moles of AlCl3 = 1.111 mol
6. Convert the moles of aluminum chloride to grams by multiplying it by the molar mass of aluminum chloride:
Mass of AlCl3 = 1.111 mol * 133.34 g/mol = 148.1 g (rounded to one decimal place)
Therefore, when 30.0g of aluminum reacts with HCl, approximately 148.1 grams of aluminum chloride are formed.