Question

How many grams of hydrochloric acid will react completely with a block of gold that is 3.2 cm by 3.8 cm by 2.8 cm, if the density of gold is 19.3 g/mL? Show all steps of your calculation as well as the final answer.

Au+HCL -> AuCl2+H2

Answer 1

243.50 grams of hydrochloric acid will react completely with a block of gold.

Step-by-step explanation:

Mass of gold = m

Length of gold block = l = 3.2 cm

Breadth of gold block = b = 3.8 cm

height of gold block = h = 2.8 cm

Volume of the gold = V = l × b × h = 3.2 cm × 3.8 × cm × 2.8 cm

V =
`34.048 cm^3=34.048 mL` (
`1 cm^3=1 mL`)

Density of gold = d = 19.3 g/ml

`m=d* V=19.3 g/ml* 34.048 mL=657.1264 g`

Moles of gold =
`(657.1264 g)/(197 g/mol)=3.3357 mol`

`Au+2HCl\rightarrow AuCl_2+H_2`

According to recation, 1 mole gold reacts with 2 moles of HCl .Then 3.3357 moles of gold will ":

`(2)/(1)* 3.3357 mol=6.6713 mol` of HCl

Mass of 6.6713 moles of HCl =

`6.6713 mol* 36.5 g/mol=243.50 g`

243.50 grams of hydrochloric acid will react completely with a block of gold.

Answer 2

1) The answer is: 243.27 grams of hydrochloric acid will react completely.

Balanced chemical reaction: Au + 2HCl → AuCl₂ + H₂.

V(Au) = 3.2 cm · 3.8 cm · 2.8 cm.

V(Au) = 34.05 cm³; volume of gold.

d(Au) = 19.3 g/cm³; density of gold.

m(Au) = V(Au) · d(Au).

m(Au) = 34.05 cm³ · 19.3 g/cm³.

m(Au) = 657.13 g; mass of gold.

n(Au) = m(Au) ÷ M(Au).

n(Au) = 657.13 g ÷ 196.97 g/mol.

n(Au) = 3.33 mol; amount of gold.

2) From balanced chemical reaction: n(Au) : n(HCl) = 1 : 2.

n(HCl) = 2 · n(Au).

n(HCl) = 2 · 3.33 mol.

n(HCl) = 6.66 mol; amount of hydrochloric acid.

M(HCl) = 36.46 g/mol; molar mass of hydrochloric acid.

m(HCl) = n(HCl) · M(HCl).

m(HCl) = 6.66 mol · 36.46 g/mol.

m(HCl) = 243.27 g; mass of hydrochloric acid.