Answer:
0.135 mole of H2.
Explanation:
We'll begin by calculating the number of mole in 3.24 g of Mg. This can be obtained as follow:
Mass of Mg = 3.24 g
Molar mass of Mg = 24 g/mol
Mole of Mg =?
Mole = mass /Molar mass
Mole of Mg = 3.24/24
Mole of Mg = 0.135 mole
Next, we shall write the balanced equation for the reaction. This is illustrated below:
Mg + 2HCl —> MgCl2 + H2
From the balanced equation above,
1 mole of Mg reacted to produce 1 mole of H2.
Finally, we shall determine the number of mole of H2 produced by reacting 3.24 g (i.e 0.135 mole) of Mg. This can be obtained as follow:
From the balanced equation above,
1 mole of Mg reacted to produce 1 mole of H2.
Therefore, 0.135 mole of Mg will also react to produce 0.135 mole of H2.
Thus, 0.135 mole of H2 can be obtained from the reaction.
Molecular mass= (14.01∗1)+(1.008∗3)
14.01+3.024=17.03g/mol
The last one 1) exothermic; 2) exothermic
I believe that would be oil and vinegar.
Answer:
First, balance the half-reactions
Second, equalize the electrons
Third,add two reaction equations to get final answer
Explanation:
For example
H₂C₂0₄ + MnO⁻₄ ---------->CO₂+Mn²⁺
(i) Balancing the half reactions
H₂C₂O₄-------->2CO₂+2H⁺+2e⁻
5e⁻ +8H⁺+MnO₄⁻----------->Mn²⁺+4H₂O
(ii)
Equalizing the electrons
5H₂C₂O₄--------->10CO₂+10H⁺+10e⁻ ---here there is a factor of 5
10e⁻+16H⁺+2MnO₄⁻--------->2Mn²⁺+8H₂O -----here there is a factor of 2
(iii)
Add the two where electrons and some Hydrogen ions will cancel out
5H₂C₂O₄+6H⁺+2MnO₄⁻---->10CO₂+2Mn²⁺+8H₂O
