Answer:
The mass of PbSO4 formed 15.163 gram
Explanation:
mole of Pb(NO₃)₂ = 1.25 x 0.05 = 0.0625
mole of Na₂SO₄ = 2 x 0.025 = 0.05
Pb(NO₃)₂ + Na₂SO₄ → PbSO₄ + 2 NaNO₃
( Mole/Stoichiometry )
= 0.0625 = 0.05
From (Mole/ Stoichiometry ) we can conclude that Na₂SO₄ is limiting reagent.
Mass of PbSO₄ precipitate = 0.05 x Molecular mass of PbSO₄
= 0.05 x 303.26 g
= 15.163 g
The equation you use here is
mass =moles x Mr
So:
Moles of K - 0.55mol
Mr of K - 39.1
Mass= 0.55x39.1 =21.505g
The reactants, products, coefficients, subscripts. ( I forgot the rest lol)
Answer:
666.5
Explanation:
Multiply 2.15 and 3.1 to get 6.665.
6.665×100≈666.5
Multiply 6.665 and 100 to get 666.5.
666.5
The chemicals are reactive with one another