91影视

a)

\(Fe(s) + SO_4^{2 - }(aq) + {H_3}{O^ + }(aq) \to F{e^{2 + }}(aq) + SO_4^{2 - }(aq) + {H_2}{O^ + }(l) + {H_2}(g)\)

balance the equation:

\(Fe(s) + SO_4^{2 - }(aq) + 2{H_3}{O^ + }(aq) \to F{e^{2 + }}(aq) + SO_4^{2 - }(aq) + 2{H_2}{O^ + }(l) + {H_2}(g)\)

b)

\(FeC{l_3}(aq) + NaOH(aq) \to Fe{(OH)_3}(s) + N{a^ + }(aq) + C{l^ - }(aq)\)

We use algebraic equation to balance the equation:

\(\begin{aligned}aFeC{l_3}(aq) + bNaOH(aq) \to cFe{(OH)_3}(s) + dN{a^ + }(aq) + eC{l^ - }(aq)\\a + b \to c + d + e\end{aligned}\)

If we balance Fe atoms, we must have,

a = c

means, a = c = 1, we get;

If we balance Cl atoms, we must have,

3a = e

And, since a = 1, we get:

e = 3

If we balance O atoms, we must have,

b = 3c

And, since c = 1, we get:

b = 3

If we balance Na atoms, we must have,

b = d

And, since b = 3, we get:

d = 3

Now, we have,

\(a + 3b \to c + 3d + 3e\)

From which we get,

\(FeC{l_3}(aq) + 3NaOH(aq) \to Fe{(OH)_3}(s) + 3N{a^ + }(aq) + 3C{l^ - }(aq)\)

c)

\(Mn{(OH)_2}(s) + HBr(aq) \to M{n^{2 + }}(aq) + B{r^ - }(aq) + {H_2}O(l)\)

We use algebraic equation to balance the equation:

\(\begin{aligned}aMn{(OH)_2}(s) + bHBr(aq) \to cM{n^{2 + }}(aq) + dB{r^ - }(aq) + e{H_2}O(l)\\a + b \to c + d + e\end{aligned}\)

If we balance Mn atoms, we must have,

a = c

means a = c = 1

If we balance O atoms, we must have,

\(2a = e\)

And, since a = 1, we get

e = 2

If we balance H atoms, we must have,

\(\begin{aligned}2a + b = 2e\\2 + b = 4\\b = 2\end{aligned}\)

If we balance Br atoms, we must have,

\(b = d\)

And, since b = 2, we get

d = 2

Now, we have,

\(a + 2b \to c + 2d + 2e\)

\(Mn{(OH)_2}(s) + 2HBr(aq) \to M{n^{2 + }}(aq) + 2B{r^ - }(aq) + 2{H_2}O(l)\)

d)

\(Cr(s) + {O_2}(g) \to C{r_2}{O_3}(s)\)

We use algebraic equation to balance the equation:

\(\begin{aligned}aCr(s) + b{O_2}(g) \to cC{r_2}{O_3}(s)\\a + b \to c\end{aligned}\)

If we balance Cr atoms, we must have,

a = 2c

If we balance O atoms, we must have,

\(\begin{aligned}2b = 3c\\b = \frac{3}{2}c\end{aligned}\)

Now we have,

\(2a + \frac{3}{2}b \to c\)

We get,

\(2Cr(s) + \frac{3}{2}{O_2}(g) \to C{r_2}{O_3}(s)\)

We need to multiply whole with 2 to get whole number,

\(4Cr(s) + 3{O_2}(g) \to 2C{r_2}{O_3}(s)\)

e)

\(M{n_2}{O_3}(s) + HCl(aq) \to MnC{l_3}(aq) + {H_2}O(l)\)

We use algebraic equation to balance the equation:

\(\begin{aligned}aM{n_2}{O_3}(s) + bHCl(aq) \to cMnC{l_3}(aq) + d{H_2}O(l)\\a + b \to c + d\end{aligned}\)

If we balance Mn atoms, we must have,

2a = c

If we balance O atoms, we must have,

\(3a = d\)

If we balance H atoms, we must have,

\(b = 2d = 2.3a = 6a\)

Now we have,

\(a + 6b \to 2c + 3d\)

We get,

\(M{n_2}{O_3}(s) + 6HCl(aq) \to 2MnC{l_3}(aq) + 3{H_2}O(l)\)

f)

\(Ti(s) + {F_2}(aq) \to Ti{F_4}(g)\)

We use algebraic equation to balance the equation:

\(\begin{aligned}aTi(s) + b{F_2}(aq) \to cTi{F_4}(g)\\a + b \to c\end{aligned}\)

If we balance Ti atoms, we must have,

a = c

means a = c = 1

If we balance F atoms, we must have,

\(2b = 4c\)

Since c = 1we get,

2b = 4

b = 2

Now we have,

\(a + 2b \to c\)

We get,

\(Ti(s) + 2{F_2}(aq) \to Ti{F_4}(g)\)