A 8.00 L tank at 26.9 C is filled with 5.53 g of dinitrogen difluoride gas and 17.3 g of sulfur hexafluoride gas. You can assume
1 answer:
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So the first thing we must do is write a balanced equation for the reaction and we know the equation is balnced when all the species on the RHS is equal to the species on the LHS
2NaOH + H₂SO₄ → Na₂SO₄<span> + 2H₂O
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So now it's time to identify what reactant you know the most for from the question (volume & conc. of H₂SO₄) and use that info to find the unknown (conc. of NaOH)
If 1000 ml of H₂SO₄ contain 0.750 mol [0.750 M is the amount of moles in
1 L (1000 ml)]
then let 15 ml of H₂SO₄ contain x mol [15 ml is the amount of the acid that took part in the reaction]
⇒ x =
= 0.01125 mol
Mole ratio of NaOH to H₂SO₄ can be obtained from the balanced equation
0 2NaOH + 1H₂SO₄ → Na₂SO₄ + 2H₂O
mole ratio of NaOH to H₂SO₄ is 2 : 1
∴ if mole of of H₂SO₄ = 0.01125 mol
then moles of NaOH = (0.01125 mol) × 2
= 0.0225 mol
If 17.5 ml of NaOH contain 0.0225 mol [this was given in the question]
then let 1000 ml of NaOH contain x
⇒ x =
= 1.286 mol
∴ concentration of NaOH is 1.286 mol/L
Answer:
0.171 grams of oxygen combined with the metal.
Explanation:
Mass of magnesium oxide produced = 0.421 g
Mass of magnesium metal used = 0.250 g
Suppose reaction completely converts the magnesium into magnesium oxide.
Mass of oxygen combined with magnesium = m
0.421 g = 0.250 g + m
m = 0.421 g - 0.250 g = 0.171 g
0.171 grams of oxygen combined with the metal.