Step 1: Idea:
Atom economy compares useful mass with total input mass. Here half of the input mass ends up in the product.
Step 2: Calculation:
Divide product weight by reactant weight: $65 \div 130 = \tfrac12$.
Convert the fraction to a percentage: $\tfrac12 \times 100 = 50\%$.
Step 3: Why the Others Fail:
If the product were 65% of the input, its weight would be $0.65\times130 = 84.5$ u. For 70% it would be 91 u and for 40% it would be 52 u. None of these equals 65 u, so only 50% fits.
Final Answer:
Atom economy equals 50 percent, so option (D).
\[ \boxed{\text{(D) } 50\%} \]