If 34.3 grams of pb02 react how many grams of water will be formed

5.15 g of water

Assuming the reaction in question is the one that occurs in lead storage batteries;

Pb + PbO₂ + 2H⁺ + 2HSO₄⁻ → 2PSO₄ + 2H₂O

The reaction is balanced.

We are given 34.3 grams of PbO₂

Required to determine the mass of water that will be formed,

We can use the following steps;

To get the number of moles we divide mass by the molar mass

Molar mass of PbO₂ = 239.2 g/mol

Therefore;

Moles = 34.3 g ÷ 239.2 g/mol

         = 0.143 moles

From the equation, 1 mole of PbO₂ reacts to produce 2 moles of water.

Therefore, 0.143 moles of PbO₂ would produce,

= 0.143 moles × 2

= 0.286 moles

To get mass we multiply number of moles by the molar mass

Molar mass of water = 18.02 g/mol

Therefore,

Mass = 0.286 moles × 18.02 g/mol

        = 5.154 g

       = 5.15 g

Therefore, 5.15 g of water will be formed

"hydrogen bonding is the attraction of h atoms in the water molecule (slight + charge) to o atoms in a neighboring water molecule (slight - charge); it is a weak interatomic force. but it is still strong enough to the water molecules "stick" together."

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer