1 mole ---------- 22.4 L at ( STP)
? moles -------- 75.8 L
75.8 x 1 / 22.4 => 3.3839 L
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Answer:
HF(aq)+NaOH(aq)→NaF(aq)+H2O(l)
Explanation:
Complete question
Dissolved hydrofluoric acid reacts with dissolved sodium hydroxide to form water and aqueous sodium fluoride. What is the net ionic equation
Equilibrium equation between the undissociated acid and the dissociated ions
HF(aq)⇌H+(aq)+F−(aq)
Sodium hydroxide will dissociate aqueous solution to produce sodium cations, Na+, and hydroxide anions, OH−
NaOH(aq)→Na+(aq)+OH−(aq)
Hydroxide anions and the hydrogen cations will neutralize each other to produce water.
H+(aq)+OH−(aq)→H2O(l)
On combining both the equation, we get –
HF(aq)+Na+(aq)+OH−(aq)→Na+(aq)+F−(aq)+H2O(l)
The Final equation is
HF(aq)+NaOH(aq)→NaF(aq)+H2O(l)
Answer:
65.08 g.
Explanation:
- For the reaction, the balanced equation is:
<em>2AlCl₃ + 3Br₂ → 2AlBr₃ + 3Cl₂,</em>
2.0 mole of AlCl₃ reacts with 3.0 mole of Br₂ to produce 2.0 mole of AlBr₃ and 3.0 mole of Cl₂.
- Firstly, we need to calculate the no. of moles of 36.2 grams of AlCl₃:
<em>n = mass/molar mass</em> = (36.2 g)/(133.34 g/mol) = <em>0.2715 mol.</em>
<u><em>Using cross multiplication:</em></u>
2.0 mole of AlCl₃ reacts with → 3.0 mole of Br₂, from the stichiometry.
0.2715 mol of AlCl₃ reacts with → ??? mole of Br₂.
∴ The no. of moles of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = (0.2715 mol)(3.0 mole)/(2.0 mole) = 0.4072 mol.
<em>∴ The mass of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = no. of moles of Br₂ x molar mass</em> = (0.4072 mol)(159.808 g/mol
) = <em>65.08 g.</em>
