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displacement reaction............ u people have to find the answers instead asking others without shame..............
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moles of Al=4.40 g/26.9815 g/mol=0.163 moles cl2=15.4g/70.906g/mol=0.217 the ratio is 2:3 cl2 is the limiting reagent
2Al(s) + 6HCl(aq) = 2AlCl3(aq) + 3H2(g). The molar ratios are 2:6 :: 2:3 This means that two moles of aluminium will react to completion with exactly six moles of hydrocchloric acid. At completion there will be two moles of aluminium chloride and three moles of hydrogen gas. So if you had 8 moles of hydrochloric acid and only 2 moles of aluminium, then the aluminium would be the limiting reactant, because from the 8 moles hydrochloric acid you are using only six moles, leaving two moies on hydrochloric acid unreacted. Conversely, if you had 3 moles of Al and 6 moles HCl , then the HCl would be the limiting reactant. Because the 6 moles of acid would only react with two moles of Al leaving one mole Al unreacted.
The reaction between NaOH (sodium hydroxide) and Al (aluminum) can be represented as: 2Al + 6NaOH → 2Na3AlO3 + 3H2. This reaction produces sodium aluminate and hydrogen gas.
2Al + 3Cl2 ---> 2AlCl3
2Al+3S--->Al2S3
Al(NO3)3 + 4NaOH -> NaAlO2 + 3NaNO3 + 2H2O
The balanced equation is: 2Al(NO3)3 + 6NaOH → 2Al(OH)3 + 6NaNO3 .
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Sure! Here are the balanced chemical equations: MnCl2 + 2Al -> 2AlCl3 + Mn 2Al + 2MnCl3 -> 2AlCl3 + 2Mn
When nitric acid reacts with aluminum, it produces aluminum nitrate, nitrogen dioxide gas, and water. The balanced chemical equation for this reaction is: 2Al + 6HNO3 -> 2Al(NO3)3 + 3H2O + 3NO2
HYDROGEN
2Al + N2 ----> 2AlN
2Al + 3CuSO4 = Al2(SO4)3 + 3Cu
2 Al + 6 H2O > 2 Al(OH)3 + 3 H2 Al(OH)3 + NaOH > Na+ + [Al(OH)4]- __________________________________________________________________ 6H2O(l)+2Al(s)+2NaOH(aq)--------------- 2NaAl(OH)4(aq)+3H2(g)