Answer:a) 0.1 mole. b) 4g. c) 2% d) 196 mL
Explanation: in 200mL , 0.1mole
mw NaOH = 40g/mol —> 4g in 0.1 mole
4g in 200mL so 2g in 100mL
density NaOH = 1g/mL so if 4g in 200 mL, 4mL , 196 mL water
<span>N2 + 3H2 → 2 </span>NH3<span> from bal. rxn., 2 moles of </span>NH3<span> are formed per 3 moles of </span>H2, 2:3 moleH2<span>: 3.64 </span>g<span>/ 2 </span>g<span>/mole </span>H2<span>= 1.82 1.82 moles </span>H2<span> x 2/3 x 17
</span>
Answer: 10.9 mol.
Explanation:
- To understand how to solve this problem, we must mention the reaction equation where water produced from PbO₂.
Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O
- Now, it is a stichiometric oriented problem, that 1 mole of PbO₂ produces 2 moles of H₂O.
Using cross multiplication:
1.0 mole of PbO₂ → 2.0 moles of H₂O
5.43 moles of PbO₂ → ??? moles of water
The moles of water produced = (5.43 x 2.0) = 10.86 moles ≅ 10.9 moles.
