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1. How many moles of sodium are required to produce 40. grams of NaOH?

2. How many grams of NaOH are produced with the reaction of 5.00 moles of water?

3. How many grams of H₂ are produced with the reaction of 36 grams of water?

User Niro
by
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2 Answers

3 votes

Answer:

1. 1 mole of Na.

2. 200g of NaOH.

3. 2g of H2.

Step-by-step explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

2Na + 2H2O —> 2NaOH + H2

1. Determination of the number of mole sodium required to produce 40 grams of NaOH.

First, we'll determine the mass of NaOH produced from the balanced equation. This is illustrated below:

Molar mass of NaOH = 23 + 16 + 1 = 40g/mol

Mass of NaOH from the balanced equation = 2 x 40 = 80g

Now, we can obtain the number of mole of Na needed to produce 40g of NaOH as follow:

From the balanced equation above,

2 moles of Na reacted to produce 80g of NaOH.

Therefore, Xmol of Na will react to produce 40g of NaOH i.e

Xmol of Na = (2 x 40)/80

Xmol of Na = 1 mole.

Therefore, 1 mole of Na is required.

2. Determination of the mass of NaOH produced by 5 moles of H2O.

This can be obtained as follow:

From the balanced equation above,

2 moles of H2O reacted to produce 80g of NaOH.

Therefore, 5 moles of H2O will react to produce = (5 x 80)/2 = 200g of NaOH.

Therefore, 200g of NaOH is produced.

3. Determination of the mass of H2 produced by the reaction of 36g of H2O.

First we'll begin by calculating the mass of H2O that reacted and the mass of H2 produced from the balanced equation. This is illustrated below:

Molar Mass of H2O = (2x1) + 16 = 18g/mol

Mass of H2O from the balanced equation = 2 x 18 = 36g

Molar Mass of H2 = 2x1 = 2g/mol.

Mass of H2 from the balanced equation = 1 x 2 = 2g

From the balanced equation above,

We can see that 36g of H2O will react to produce 2g of H2.

Therefore, 2g if H2 is produced by the reaction.

User Marjani
by
6.4k points
3 votes

Answer:

1.
n_(Na)=1molNa

2.
m_(NaOH)=200gNaOH

3.
m_(H_2)=2gH_2

Step-by-step explanation:

Hello,

In this case, the given chemical reaction is:


2Na+2H_2O\rightarrow 2NaOH+H_2

Thus:

1. since sodium (23 g/mol) and sodium hydroxide (40 g/mol) are in a 2:2 molar ratio we compute the required moles of sodium:


n_(Na)=40gNaOH*(1molNaOH)/(40gNaOH)*(2molNa)/(2molNaOH) =1molNa

2. Since water and sodium hydroxide are also in a 2:2 mole ratio, we compute the grams:


m_(NaOH)=5.00molH_2O*(2molNaOH)/(2molH_2O)*(40gNaOH)/(1molNaOH) =200gNaOH

3. Now, we notice water (18 g/mol) and hydrogen (2 g/mol) are in a 2:1 molar ratio, so we proceed as follows:


m_(H_2)=36gH_2O*(1molH_2O)/(18gH_2O) *(1molH_2)/(2molH_2O)*(2gH_2)/(1molH_2) =2gH_2

Regards.

User Sheldonk
by
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