Using this balanced equation: 2 NaOH + H2SO4 —> H2O + Na2SO4

How many grams of sodium sulfate will be formed if you start with 150 grams of
sodium hydroxide and you have an excess of sulfuric acid?

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Respuesta :

Answer:

266. 25 g

Explanation:

Data given:

Sodium hydroxide (NaOH) = 150 g

Sodium sulfate Na₂SO₄= ?

Reaction Given:

                       2 NaOH + H₂SO₄—> H₂O + Na₂SO₄

Solution:

First we have to look at the reaction

                       2 NaOH + H₂SO₄—> H₂O + Na₂SO₄

                       2 mol                                    1 mol

So,

2 mole sodium hydroxide gives 1 mole Na₂SO₄

Now,

Convert moles to mass

for this purpose we have to know molar mass of sodium hydroxide (NaOH) and sodium sulfate (Na₂SO₄)

Molar mass of NaOH = 23 + 16 +1

Molar mass of NaOH = 40 g

Molar mass of Na₂SO₄ = 2(23) + 32+ 4(16)

Molar mass of Na₂SO₄ = 46 + 32+ 64

Molar mass of Na₂SO₄ = 142

Mole Mass Conversion:

                 2 NaOH        +          H₂SO₄  —>   H₂O   +   Na₂SO₄

                 2 mol (40g/mol)                                       1 mol (142 g/mol)

                       

So

                  2 NaOH + H₂SO₄—> H₂O + Na₂SO₄

                  80 g                                        142 g

Apply unity formula

                      80 g of NaOH ≅ 142 g of Na₂SO₄

                      150 g of NaOH ≅ X g of Na₂SO₄

by Cross multiplication

                  X g of Na₂SO₄ = 142g x 150g / 80 g

                  X g of Na₂SO₄ = 266. 25 g

So 150 g of sodium hydroxide will give 266.25 g Na₂SO₄