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How many grams of water are needed to react with 18.9 grams of li2o? given: li2o+h2oâ2lioh?

4vir4ik [10]

The balanced equation for the reaction is as follows;
Li₂O + H₂O ---> 2LiOH
Stoichiometry of Li₂O to H₂O is 1:1
Mass of Li₂O reacted - 18.9 g
Number of Li₂O moles reacted - 18.9 g / 30 g/mol = 0.63 mol
An equivalent amount of moles of water have reacted - 0.63 mol
mass of water required - 0.63 mol x 18 g/mol = 11.34 g
A mass of 11.34 g of water is required

4 0

3 years ago

How many seconds are in 5 hours

vfiekz [6]

18000 seconds = 5 hours

8 0

3 years ago

Read 2 more answers

When the equation Sn + HNO₃ → SnO₂ + NO₂ + H₂O is balanced in acidic solution, what is the smallest whole-number coefficient for

balu736 [363]

Answer:

1.

Explanation:

Hello,

In this case, for the given reaction we first assign the oxidation state for each species:

$Sn^0 + H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 + N^{4+}O^{2-}_2 + H^+_2O^-$

Whereas the half reactions are:

$Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow N^{4+}O^{2-}_2+H_2O$

Next, we exchange the transferred electrons:

$1\times(Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-)\\\\4\times (H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow N^{4+}O^{2-}_2+H_2O)\\\\\\Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\4H^++4H^+N^{5+}O^{-2}_3 +4e^-\rightarrow 4N^{4+}O^{2-}_2+4H_2O$

Afterwards, we add them to obtain:

$Sn^0+2H_2O+4H^++4H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 +4H^++4N^{4+}O^{2-}_2+4H_2O$

By adding and subtracting common terms we obtain:

$Sn^0+4H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 +4N^{4+}O^{2-}_2+2H_2O$

Finally, by removing the oxidation states we have:

$Sn + 4HNO_3 \rightarrow SnO_2 + 4NO_2 + 2H_2O$

Therefore, the smallest whole-number coefficient for Sn is 1.

Regards.

6 0

3 years ago

Turbines are 100 percent efficient. True False

max2010maxim [7]

False because there’s only 60% turbulence 20% heat 20% sound

6 0

3 years ago

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Calculate the standard cell potential given the following standard reduction potentials: Al3++3e−→Al;E∘=−1.66 V Cu2++2e−→Cu;E∘=0

Maru [420]

Answer:

Explanation:

To calculate the cell potential we use the relation:

Eº cell = Eº oxidation + Eº reduction

Now in order to determine which of the species is going to be oxidized, we have to remember that the more the value of the reduction potential is negative, the greater its tendency to be oxidized is. In electrochemistry we use the values of the reductions potential in the tables for simplicity because the only thing we need to do is change the sign of the reduction potential for the oxized species .

So the species that is going to be oxidized is the Aluminium, and therefore:

Eº cell = -( -1.66 V ) + 0.340 V = 5.06 V

Equally valid is to write the equation as:

Eº cell = Eº reduction for the reduced species - Eº reduction for the oxidized species

These two expressions are equivalent, choose the one you fell more comfortable but be careful with the signs.

3 0

2 years ago