4HCl + O₂ → 2Cl₂ + 2H₂O
mole ratio of HCl : O₂ is 4 : 1
∴ if moles of HCl = 2.3 mol
then mol of O₂ = 2.3 mol ÷ 4
= 0.575 mol
mass of O₂ = moles of O₂ × molar mass of O₂
= 0.575 mol × (16 × 2) g/mol
= 18.4 g
Answer: B, the gas is being squeezed out of the liquid.
Explanation: The gas does not want to be trapped inside of the liquid, so it is trying to force it’s way out. Therefore creating more pressure in the container or whatever the liquid is being held in.
Answer:
d) 0.1202 M
Explanation:
Let's consider the neutralization reaction between NaOH and a generic monoprotic acid.
NaOH + HA → NaA + H₂O
The used volume of NaOH is 41.63 mL - 19.63 mL = 22.00 mL. The moles of NaOH are:
22.00 × 10⁻³ L × 0.1093 mol/L = 2.405 × 10⁻³ mol
The molar ratio of NaOH to HA is 1:1. The moles of HA that reacted are 2.405 × 10⁻³ moles.
The molar concentration of HA is:
2.405 × 10⁻³ mol / 20.00 × 10⁻³ L = 0.1202 M
Fe2(SO4)3.3H2O = 56*2 + 3*(32+16*4)+3*(1*2+16) = 454 g/mol
Mol =m/M = 10.5/454 = 0.023 mol
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Hg(No3)2 +NaSO4 --->2NaNO3 + HgSO4(s)
calculate the moles of each reactant
moles=mass/molar mass
moles of Hg(NO3)2= 51.429g/ 324.6 g/mol(molar mass of Hg(NO3)2)=0.158 moles
moles Na2SO4 16.642g/142g/mol= 0.117 moles of Na2SO4
Na2SO4 is the limiting reagent in the equation and by use mole ratio Na2So4 to HgSO4 is 1:1 therefore the moles of HgSO4 =0.117 moles
mass of HgSO4=moles x molar mass of HgSo4= 0.117 g x 303.6g/mol= 35.5212 grams