Question

How many atoms of mercury are required to produce 3.49 L of H2 gas (density =
0.0899 g/L
according to the balanced chemical reaction:
2Hg(l) + 2 HNO3(aq) → Hg2(NO3)2(aq) + H2(g)

Answer 1

The number of mercury atoms required to produce 3.49 L of H2 gas according to the equation would be 1.89 x $10^{23}$

From the balanced equation of the reaction:

$2Hg(l) + 2 HNO_3(aq) ---> Hg_2(NO_3)_2(aq) + H_2(g)$

2 moles of Hg are required to produce 1 mole of the H2 gas.

3.49 L of H2 gas was produced and the density of the gas is 0.0899 g/L

Recall that: density = mass/volume

Mass of the H2 gas = density x volume

                                     = 3.49 x 0.0899

                                     = 0.3138 g of H2 gas

Mole of 0.3138 g of H2 gas = mass/molar mass

                                         = 0.3138/2

                                            = 0.1569 mole

Since the ratio of Hg to H2 according to the equation is 2:1, it means that the mole of Hg is twice the mole of H2.

Thus, mole of Hg = 0.1569 x 2

                                    = 0.3138 mole

According to Avogadro, 1 mole of a substance = 6.022×10^23

0.3138 mole would be:

                 0.3138 x 6.022 x 10^23

                            = 1.89 x $10^{23}$

More on the number of atoms in molecules can be found here: https://brainly.com/question/19036641