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
