Question

sodium oxide (na₂o) can react with hydrochloric acid (hcl) to produce sodium chloride (nacl) and water (h₂o) according to the following equation: na₂o + 2hcl -> 2nacl + h₂o. how many total atoms are present before and after the reaction? a 2 before and 2 after b 3 before and 3 after c 5 before and 5 after d 7 before and 7 after

Explanation:

Step1: Count atoms before reaction

In $Na_2O$, there are 2 sodium (Na) and 1 oxygen (O) atoms. In $2HCl$, there are 2 hydrogen (H) and 2 chlorine (Cl) atoms. So in total, $2 + 1+2 + 2=7$ atoms before reaction.

Step2: Count atoms after reaction

In $2NaCl$, there are 2 sodium (Na) and 2 chlorine (Cl) atoms. In $H_2O$, there are 2 hydrogen (H) and 1 oxygen (O) atoms. So in total, $2+2 + 2+1 = 7$ atoms after reaction.

Answer:

D. 7 before and 7 after