2H2SO4+ Pb(OH)4 = Pb(SO4)+ r H2O

1) What is the mole ratio between H2SO4 and Pb(OH)4?

2) How many grams of H2SO4 would be required to react completely with 15.00 grams of Pb(OH)4?

3) If 25 miles of H2SO4 reacted, how many moles of water would be produced?

4) If 65.0 grams of H2SO4 reacts then, how many grams of water would be formed?

Can someone answer these questions please

Explanation:

2H2SO4 + Pb(OH)4 → 4 H2O + Pb(SO4)2

1. Mole Ratio

From the balanced equation, 2 moles of H2SO4 reacted with 1 mole of Pb(OH)4. The mole ration is therefore 2:1

2. 1 mole of H2SO4 reacts completely with 1 mole of Pb(OH)4

A mole contains the molar mass

1 mole of H2SO4 = 98 g/mol

1 mole of  Pb(OH)4 = 275 g/mol

98g of H2SO4 reacts completely with 275g of Pb(OH)4

xg would react with  15g of Pb(OH)4

x = 15 * 98 / 275

x = 5.35g

3. From the balanced equation, 2 mole of H2SO4 produces 4 moles oof H2O.

The mole ratio is 2:4 = 1; 2

This means 25 moles of H2SO4 would produce x moles of H2O

x = 25 * 2 = 50 moles of H2O

3. 2 moles of H2SO4 produces 4 moles of H2O. in terms of masses we have;

(2 * 98)g of H2SO4 produces (4 * 18)g of H2O

65g would produce xg of H2O

x = 65 * 72 / 196 = 23.88g of H2O

not clear