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6. using the following equation: 2 naohh2so4 --> 2 h2o+na2so4
how many grams of sodium sulfate will be formed if you start with 200 grams of sodium hydroxide and you have an excess of sulfuric acid?

7. using the following equation: pb(so4)2+4 lino3 --> pb(no3)4+2 liso4
how many grams of lithium nitrate will be needed to make 250 grams of lithium sulfate, assuming that you have an adequate amount of lead (iv) sulfate to do the reaction?

6. 355.1 g of Na₂SO₄ can be formed.

7. 313 g of LiNO₃ were needed

Explanation:

Excersise 6.

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

Excersise 7.

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