Answer:
Explanation:
complete combustion reaction of ethane is given by the reaction
2C2H6+7O2..............4CO2+6H2O
no of moles in 34 grams of O2=34/32=1.063
7mole of O2 produced 6 moles of H2O
therefore 1.063 moles of O2 produced=1.063*6/7=0.9 moles
now 0.9 moles of H2O contain how much grams=0.9*18=16.2 grams
Answer:
maybe 200g
I don't know thats just a guess
HOPE ITS TURE
Volume can be measures in liters
Crystal<span> of tetrahedrally bonded carbon atoms in a </span>covalent network lattice<span> witch </span>crystallizes<span> into the diamond </span>lattice.
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</span>
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 molMole 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 molIf 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