Answer:
48.84mL
Explanation:
H2SO4 + 2KOH → K2SO4 + 2H2O
From the question:
nA = 1
nB = 2
From the question given we obtained the following information:
Ma = 0.43M
Va =?
Mb = 0.35M
Vb = 120mL
Using MaVa / MbVb = nA/nB, we can easily find the volume of the acid required. This is illustrated below:
MaVa / MbVb = nA/nB
0.43 x Va / 0.35 x 120 = 1/2
Cross multiply to express in linear form
2 x 0.43 x Va = 0.35 x 120
Divide both side by the (2 x 0.43)
Va = (0.35 x 120) /(2 x 0.43)
Va = 48.84mL
Therefore, the volume of H2SO4 required is 48.84mL
C. N₂ + 3H₂ → 2NH₃
In a redox, or oxidation-reduction, reaction, one of the reactants must be reduced, which means its oxidation number decreases, while the other reactant must be oxidized, which means its oxidation number increases. The oxidation number of nitrogen in the reactants is 0 and is -3 in the products, so it is reduced. Similarly, the oxidation number of hydrogen is 0 in the reactants but it is +1 in the products.
Balanced chemical reaction happening here is:
3Mg(s) + N₂(g) → Mg₃N₂(s)
<u>moles of product formed from each reactant:</u>
2.0 mol of N2 (g) x <u> 1 mol Mg₃N₂ </u> = <u>2 mol Mg₃N₂</u>
1 mol N2
and
8.0 mol of Mg(s) x <u> 1 mol Mg₃N₂ </u> = 2.67 mol Mg₃N₂
3 mol Mg
Since N2 is giving the least amount of product(Mg₃N₂) ie. 2 mol Mg₃N₂
N2 is the limiting reactant here and Mg is excess reactant.
Hence mole of product formed here is 2 mol Mg₃N₂
molar mass of Mg₃N₂
= 3 Mg + 2 N
= 101g/mol
mass of product(Mg₃N₂) formed
= moles x Molar mass
= 2 x 101
= 202g Mg₃N₂
<u>202g of product are formed from 2.0 mol of N2(g) and 8.0 mol of Mg(s).</u>
<u> </u> The following are indicators of chemical changes:
Change in Temperature
Change in Color
Formation of a Precipitate