Convert the child weight (37.3 pounds) to kilograms
37.3 lb x 0.453 kg /1lb = "A kg"
multiply the dose (9.00mg/kg) by the weight of the child to find how much you need to give him
A kg * 9.00 mg/1kg = "B mg"
calculate the mL of suspension dividing the "B mg" by the concentration of the suspension 60.0 mg/mL
B mg * 1mL/ 60.0 mg = C mL <span>oxcarbazepine</span>
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.
The answer lies in the stoichiometry of the reaction. If u look at the number BEFORE the reagent u will see the ratios of the reagents.
D is the answer since it is changing the element.
