Multiple choice

If $0.5$ g of a mixture of two metals, $A$ and $B$ with respective equivalent weights, $12$ and $9$ displaces $560$ mL of $H$ at STP from an acid, the composition of the mixture is :

  1. $40\% A, 60\% B$
  2. $60\% A, 40\% B$
  3. $30\% A, 70\% B$
  4. $70\% A, 30\% B$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

560 mL of H2 at STP is 0.025 moles, which corresponds to 0.05 equivalents. Let mass of A be x, then mass of B is 0.5-x. Equivalents equation: x/12 + (0.5-x)/9 = 0.05. Solving gives x = 0.2, so A is 0.2/0.5 = 40% and B is 60%.

AI explanation

The 560 ml of hydrogen at STP corresponds to 0.025 moles, which means the mixture produces 0.025 gram equivalents of hydrogen. Using the equivalent weight concept, we set up the equation x divided by 12 plus 0.5 minus x divided by 9 equals 0.025 to find the masses of the metals. Solving this equation gives the mass of metal A as 0.2 grams and the mass of metal B as 0.3 grams. The composition is 0.2 divided by 0.5, giving 40 percent for metal A, and 0.3 divided by 0.5, giving 60 percent for metal B. The composition of the mixture is 40 percent A and 60 percent B.

Keep practicing — related questions