Under the right conditions aluminum will react with chlorine to produce aluminum chloride. 2 Al + 3Cl2 -> 2 AlCl3
How many grams of aluminum chloride can be produced when 157.0g of aluminum react with excess chlorine?
775.9 g AlCl3
but what the steps pls help

775.9 g

Explanation:

Step 1: Write the balanced equation

2 Al + 3 Cl₂ ⇒ 2 AlCl₃

Step 2: Calculate the mass corresponding to 157.0 g of Al

The molar mass of Al is 26.98 g/mol.

157.0 g × 1 mol/26.98 g = 5.819 mol

Step 3: Calculate the moles of AlCl₃ produced from 5.819 moles of Al

The molar ratio of Al to AlCl₃ is 2:2. The moles of AlCl₃ produced are 2/2 × 5.819 mol = 5.819 mol.

Step 4: Calculate the mass corresponding to 5.819 moles of AlCl₃

The molar mass of AlCl₃ is 133.34 g/mol.

5.819 mol × 133.34 g = 775.9 g

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally