How much aluminum nitrate can be found when 21 g of aluminum react with lead(II) nitrate? Answer in units of mol.?
回答 (2)
2018-01-20 9:25 am
Molar mass of Al = 27.0 g/mol
No. of moles of Al reacted = (21 g) / (27.0 g/mol) = 7/9 mol
2Al + 3Pb(NO₃)₂ → 2Al(NO₃)₃ + 3Pb
Mole ratio Al : Al(NO₃)₃ = 2 : 2 = 1 : 1
No . of moles of Al(NO₃)₃ formed = 7/9 mol = 0.78 mol (to 2 sig. fig.)
No. of moles of Al reacted = (21 g) / (27.0 g/mol) = 7/9 mol
2Al + 3Pb(NO₃)₂ → 2Al(NO₃)₃ + 3Pb
Mole ratio Al : Al(NO₃)₃ = 2 : 2 = 1 : 1
No . of moles of Al(NO₃)₃ formed = 7/9 mol = 0.78 mol (to 2 sig. fig.)
2018-01-20 5:21 pm
the reaction
.. 2 Al + 3 Pb(NO3)2 ----> 2 Al(NO3)3 + 3 Pb
via that dimensional analysis you've been studying
.. 21g Al .. 1 mol Al... . 2 mol Al(NO3)2
----- ---- x ---- ---- ---- x --- ---- ----- ----- = 0.78 mol Al(NO3)2
.. .. .1.... .. 26.98g Al.. ... 2 mol Al
.. 2 Al + 3 Pb(NO3)2 ----> 2 Al(NO3)3 + 3 Pb
via that dimensional analysis you've been studying
.. 21g Al .. 1 mol Al... . 2 mol Al(NO3)2
----- ---- x ---- ---- ---- x --- ---- ----- ----- = 0.78 mol Al(NO3)2
.. .. .1.... .. 26.98g Al.. ... 2 mol Al
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