From 5 L to moles, just divide 5 by 22.4. I got 0.22 moles of H2.
From 5 moles to liters, just multiply 5 by 22.4. I got 112 L of H2.
Answer:
[KOH] = 0.10M in KOH
Explanation:
Molar Concentration [M] = moles solute/volume solution in liters
moles KOH = 0.56g/56g/mole = 0.01mole
Volume of solution = 100cm³ = 100ml = 0.10 liter
[KOH] = 0.01 mole KOH / 0.10 liter solution = 0.10M in KOH
Answer:
AsF3:C2CI6
4:3
1.3618 moles: 1.02135 moles(1.3618÷4×3)
C2CI6 is the limting reagent
So the number of moles for AsCI3 is 0.817 moles( number of moles of the limting reagant) ÷3 ×4 (according to ratio by balancing chemical equation)=1.09 moles(3 s.f.)
or
Balanced equation
4AsF3 + 3C2Cl6 → 4AsCl3 + 3C2Cl2F4
Use stoichiometry to calculate the moles of AsCl3 that can be produced by each reactant.
Multiply the moles of each reactant by the mole ratio between it and AsCl3 in the balanced equation, so that the moles of the reactant cancel, leaving moles of AsCl3.
Explanation:
Based on the balanced chemical reaction presented above, every mole of magnesium (Mg) yields one mole of diatomic hydrogen (H2). When converted to masses, every 24.3 grams of magnesium yields 2 grams of hydrogen.
From the given, there are 20 grams of magnesium available for the reaction. With this amount, the expected yield of hydrogen is 1.646 grams. To calculate the percent yield, divide the actual yield to the hypothetical yield.
*The case is impossible because the actual yield is greater than the theoretical yield.
If we assume that there had been a typographical error and that the actual yield is 0.7 grams instead of 1.7 grams, the percent yield becomes 42.5%. Thus, the answer is letter E.
