Answer:
6. 355.1 g of Na₂SO₄ can be formed.
7. 313 g of LiNO₃ were needed
Explanation:
<u>Excersise 6</u>.
The reaction is: 2 NaOH + H₂SO₄ --> 2 H₂O + Na₂SO₄
2 moles of sodium hydroxide react with 1 mol of sulfuric acid to produce 2 moles of water and 1 mol of sodium sulfate.
If we were noticed that the acid is in excess, we assume the NaOH as the limiting reactant. Let's convert the mass to moles (mass / molar mass)
200 g / 40 g/mol = 5 moles.
Now we apply a rule of three with the ratio in the reaction, 2:1
2 moles of NaOH produce 1 mol of sodium sulfate.
5 moles of NaOH would produce (5 .1)/2 = 2.5 moles
Let's convert these moles to mass (mol . molar mass)
2.5 mol . 142.06 g/mol = 355.1 g
<u>Excersise 7.</u>
The reaction is:
Pb(SO₄)₂+ 4 LiNO₃ → Pb(NO₃)₄ + 2Li₂SO₄
As we assume that we have an adequate amount of lead (IV) sulfate, the limiting reactant is the lithium nitrate.
Let's convert the mass to moles (mass / molar mass)
250 g / 109.94 g/mol = 2.27 moles
Let's make a rule of three. Ratio is 2:4.
2 moles of lithium sulfate were produced by 4 moles of lithium nitrate
2.27 moles of Li₂SO₄ would have been produced by ( 2.27 .4) / 2 = 4.54 moles.
Let's convert these moles to mass (mol . molar mass)
4.54 mol . 68.94 g/mol = 313 g
). This is because the oxidation number of pure elements (whether alone as an atom or combined with other atoms of the same element such as the diatomic nitrogen) is always zero. 
) compounds where nitrogen has an oxidation number of zero.