Answer:
[tex]n_{PbCl_4}=0.283molPbCl_4[/tex]
Explanation:
Hello,
In this case, since lead (IV) chloride is in a 1:2 molar ratio with calcium hydroxide due to the following chemical reaction:
[tex]PbCl_4 (aq) + 2 Ca(OH)_2\rightarrow Pb(OH)_4+2CaCl_2[/tex]
We can easily compute the moles of lead (IV) chloride that will be completely consumed by 42.0 grams of calcium hydroxide whose molar mass is 74.093 g/mol:
[tex]n_{PbCl_4}=42.0gCa(OH)_2*\frac{1molCa(OH)_2}{74.093 gCa(OH)_2}*\frac{1molPbCl_4}{2molCa(OH)_2} \\\\n_{PbCl_4}=0.283molPbCl_4[/tex]
Best regards.
Answer:
0.283 moles of lead (IV) chloride are needed to completely react with 42.0 grams of calcium hydroxide.
Explanation:
First of all you must know the amount of mass of calcium hydroxide that reacts by stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction). For that you must know the molar mass of the compound. So, being:
the molar mass of calcium hydroxide is:
Ca(OH)₂= 40 g/mole + 2*(16 g/mole + 1 g/mole)= 74 g/mole
If 2 moles of sodium hydroxide react with stoichiometry, then:
Ca(OH)₂=2 moles* 74 g/mole= 148 g
Now you apply the following rule of three: if 148 grams of calcium hydroxide react with stoichiometry with 1 mole of lead (IV) chloride, how many moles of lead (IV) chloride would react with 42 grams of the hydroxide?
[tex]moles of lead (IV) chloride=\frac{42gramsof calcium hidroxide*1 mole of lead(IV) chloride}{148gramsof calcium hidroxide}[/tex]
moles of lead (IV) chloride=0.283 moles
0.283 moles of lead (IV) chloride are needed to completely react with 42.0 grams of calcium hydroxide.
