Chemistry, 06.05.2020 07:29 jalexyinez

In an experiment similar to the zinc chloride experiment, a student placed a piece of lead in hydrochloric acid. Hydrogen gas was given off, and then the liquid was boiled off. The remaining solid, lead chloride was massed. Use the data below to determine the empirical formula of lead chloride.

mass of beaker 204.35 g
mass of lead and beaker before reaction 214.71 g
mass of lead chloride and beaker after reaction 218.26 g

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Answer from: jhuss03

The empirical formula for the given compound is PbCl_2

Explanation:

We are given:

Mass of beaker, m_1 = 204.35 g

Mass of beaker and lead before the reaction, m_2 = 214.71 g

Mass of beaker and lead chloride after the reaction, m_3 = 218.26 g

Mass of lead = m_2-m_1 = [214.71 - 204.35] = 10.36 g

Mass of chlorine = m_3-m_2 = [218.26 - 214.71] = 3.55 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Lead = $\frac{\text{Given mass of Lead}}{\text{Molar mass of Lead}}=\frac{10.36g}{207.2g/mole}=0.05moles$

Moles of Chlorine = $\frac{\text{Given mass of Chlorine}}{\text{Molar mass of Chlorine}}=\frac{3.55g}{35.5g/mole}=0.1moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.05 moles.

For Lead = $\frac{0.05}{0.05}=1$

For Chlorine = $\frac{0.1}{0.05}=2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of Pb : Cl = 1 : 2

Hence, the empirical formula for the given compound is PbCl_2

Answer from: Quest

you need to take a look at the periodic system in order to sort these elements into pairs that would most likely exhibit similar chemical properties. first, you need to find elements that are found in the same group (groups are vertical in the periodic system), and then to see whether they have the same valence number. if they meet these criteria, they will have similar properties.

the first pair: ne (neon) and kr (krypton), are noble gases found in group 8

the second pair: li (lithium) and k (potassium), found in group 1

the third pair: s (sulfur) and te (tellurium), found in group 6

Answer from: Quest

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2

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In an experiment similar to the zinc chloride experiment, a student placed a piece of lead in hydroc...