Question

How many moles of lead (IV) chloride are needed to completely react with

42.0 grams of calcium hydroxide in the reaction below?

1 PbCl4 (aq) + 2 Ca(OH)2 (5)

O0.283 mol

114 mol

3.53 mol

1560 mol​

Answer 1

0.283 moles of lead (IV) chloride are needed to completely react with 42.0 grams of calcium hydroxide.

Step-by-step 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:

• Ca: 40 g/mole
• O: 16 g/mole
• H: 1 g/mole

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?

`moles of lead (IV) chloride=(42gramsof calcium hidroxide*1 mole of lead(IV) chloride)/(148gramsof calcium hidroxide)`

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.

Answer 2

`n_(PbCl_4)=0.283molPbCl_4`

Step-by-step 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:

`PbCl_4 (aq) + 2 Ca(OH)_2\rightarrow Pb(OH)_4+2CaCl_2`

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:

`n_(PbCl_4)=42.0gCa(OH)_2*(1molCa(OH)_2)/(74.093 gCa(OH)_2)*(1molPbCl_4)/(2molCa(OH)_2) \\\\n_(PbCl_4)=0.283molPbCl_4`

Best regards.