Answer:
514.5 g.
Explanation:
- The balanced equation of the reaction is: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O.
- It is clear that every 2.0 moles of NaOH react with 1.0 mole of H₂SO₄ to produce 1.0 mole of Na₂SO₄ and 2.0 moles of 2H₂O.
- Since NaOH is in excess, so H₂SO₄ is the limiting reactant.
- We need to calculate the no. of moles of 355.0 g of H₂SO₄:
n of H₂SO₄ = mass/molar mass = (355.0 g)/(98.0 g/mol) = 3.622 mol.
Using cross multiplication:
∵ 1.0 mol H₂SO₄ produces → 1.0 mol of Na₂SO₄.
∴ 3.622 mol H₂SO₄ produces → 3.662 mol of Na₂SO₄.
- Now, we can get the theoretical mass of Na₂SO₄:
∴ mass of Na₂SO₄ = no. of moles x molar mass = (3.662 mol)(142.04 g/mol) = 514.5 g.
Answer:2817.8
Explanation:multiply the value by 365
1. The molar mass of Fe2(CO3)3 is 291.72 g/mol. This means that 45.6 g is equivalent to 0.156 mol. Dividing by the 0.167 L of water gives a solution of 0.936 M.
2. Multiplying (0.672 M)(0.025 L) = 0.0168 mol. The molar mass of Ni(OH)2 is 92.71 g/mol, so multiplying by 0.0168 mol = 1.56 grams. Therefore you would need to dissolved 1.56 g of Ni(OH)2 into 25 mL of water.
3. Fe2(CO3)3 + Ni(OH)2 --> Fe(OH)3 + NiCO3Balancing: Fe2(CO3)3 + 3Ni(OH)2 --> 2Fe(OH)3 + 3NiCO3The reaction quotient is:[Fe(OH)3]^2 * [NiCO3]^3 / [Fe2(CO3)3][Ni(OH)2]^3= (0.05)^2 * (1.45)^3 / (0.936)(0.672)^3= 0.0268Since this is < 1, it implies that the reactants are favored at equilibrium.
Answer:
option B
cementation and compaction
