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

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:

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:

Best regards.

- calcium has greater affinity to bond with electron receptors than hydrogen does and hence, it has the ability to replace hydrogen in most of its compounds.

- lithium has 2 valence electrons in the product because it can make only 1 bond with hydrogen. beryllium has 4 valence electrons in the product because it can only make 2 bonds. it has fewer electrons than neon and can react with electron donors. each hydrogen has 2 valence electrons, the same as helium.

- the covalent bond formed between carbon and hydrogen is polar because carbon is more electronegative than hydrogen and it attracts the bonded electron pair closer to itself, thus giving hydrogen a partial positive charge.

- a hydrogen bond is a relatively weak bond that hydrogen atoms make with the electronegative atoms nitrogen, oxygen or fluorine.

so the best choice is a.) 2 calcium (ca) atoms

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :