Answer:
It has 1 carbon atom and 4 hydrogen atoms
Explanation:
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Answer:
HCl acts as an excess reagent
Explanation:
The reaction equation is;
PbO + 2HCl → PbCl2 + H20
Number of moles of PbO = 13g/223.2 g/mol = 0.058 moles
Since the mole ratio is 1:1, 0.058 moles of PbCl2 is produced
Number of moles of HCl = 6.4 g/36.5g/mol = 0.175 moles
2 moles of HCl yields 1 mole of PbCl2
0.175 moles yields 0.175 moles * 1/2
= 0.0875 moles of PbCl2
Hence; PbO yields the least number of moles of product so it is the limiting reactant and HCl is the reactant in excess
Answer:
2.56 grams of H₂S is needed to produce 18.00g of PbS if the H2S is reacted with an excess (unlimited) supply of Pb(CH₃COO)₂
Explanation:
The balanced reaction is:
Pb(CH₃COO)₂ + H₂S → 2 CH₃COOH + PbS
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) they react and produce:
- Pb(CH₃COO)₂: 1 mole
- H₂S: 1 mole
- CH₃COOH: 2 moles
- PbS: 1 mole
In this case, to know how many grams of H₂S are needed to produce 18.00 g of PbS, it is first necessary to know the molar mass of the compounds H₂S and PbS and then to know how much it reacts by stoichiometry. Being:
- H: 1 g/mole
- S: 32 g/mole
- Pb: 207 g/mole
The molar mass of the compounds are:
- H₂S: 2* 1 g/mole + 32 g/mole= 34 g/mole
- PbS: 207 g/mole + 32 g/mole= 239 g/mole
So, by stoichiometry they react and are produced:
- H₂S: 1 mole* 34 g/mole= 34 g
- PbS: 1 mole* 239 g/mole= 239 g
Then the following rule of three can be applied: if 239 grams of PbS are produced by stoichiometry from 34 grams of H₂S, 18 grams of PbS from how much mass of H₂S is produced?
mass of H₂S= 2.56 grams
<u><em>2.56 grams of H₂S is needed to produce 18.00g of PbS if the H2S is reacted with an excess (unlimited) supply of Pb(CH₃COO)₂</em></u>
The Boyle-Mariotte's law or Boyle's law is one of the laws of gases that <u>relates the volume (V) and pressure (P) of a certain amount of gas maintained at constant temperature</u>, as follows:
PV = k
where k is a constant.
We can relate the state of a gas at a specific pressure and volume to another state in which the same gas is at different P and V since the product of both variables is equal to a constant, according to the Boyle's law, which will be the same regardless of the state of the gas. In this way,
P₁V₁ = P₂V₂
Where P₁ and V₁ is the pressure and volume of the gas to a state 1 and P₂ and V₂ is the pressure and volume of the same gas in a state 2.
In this case, in the state 1 the gas occupies a volume V₁ = 100 mL at a pressure of P₁ = 150 kPa. Then, in the state 2 the gas occupies a volume V₂ (that we must calculate through the boyle's law) at a pressure of P₂ = 200 kPa. Substituting these values in the previous equation and clearing V₂, we have,
P₁V₁ = P₂V₂ → V₂ =
→ V₂ =
→ V₂ = 75 mL
Then, the volume occupied by the gas at 200 kPa is V₂ = 75 mL