Question

Calculate the number of hydrogen atoms in 9.60 g of (NH₄)₂CO₃ compound​

Answer 1

Answer:

The number of hydrogen atoms in 9.60 g of $(NH_{4} )_{2} CO_{3}$ is $4.82*10^{23}$ atoms.

Explanation:

Given mass = 9.60 g

Molar mass of $(NH_{4} )_{2} CO_{3}$ = (14 * 2) + (1 * 8) + (12 * 1) + (16 * 3)

                                           = 96 g/mol

Number of moles = given mass/molar mass

                             = 9.60/96

                             = 0.1 mol

1 mole $(NH_{4} )_{2} CO_{3}$ contains $6.022 *10^{23}$ molecules

0.1 mole $(NH_{4} )_{2} CO_{3}$ contains $0.1*6.022*10^{23}$ molecules

1 mole $(NH_{4} )_{2} CO_{3}$ contains 8 hydrogen atoms

0.1 mole $(NH_{4} )_{2} CO_{3}$ contains $8*0.1*6.022*10^{23}$

                                                 = $4.82*10^{23}$ atoms