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Semmy [17]
3 years ago
8

How many moles of H2O can you make from 16 moles of NaOH

Chemistry
1 answer:
dusya [7]3 years ago
5 0

Answer: 1.     For the reaction: 2K + 0.5 O2 à 1K2O

a. How many moles of O2 are needed to react with 0.56 moles of K?

0.56 moles of K  

b. How many moles of O2 are needed to make 7.6 g of K2O?

7.6 g K2O (mole/94g K2O) = 0.0809 moles K2O

0.0809 moles K2O

c. How many grams of K2O will be produced from 0.50 g of K?

0.50 g K (mole/39g K) = 0.0128 moles K

0.0128 moles K

x = 0.0064 moles K2O (78+16)g/mole =  0.60 g

2.     For the reaction: Na2O + H2O à2 NaOH

a. What mass NaOH could be made from 12.4 g of Na2O?

12.4 g of Na2O (mole/62 g) = 0.2 mole Na2O

0.2 mole Na2O

0.4 moles NaOH(40 g/mole) = 16 g NaOH

b. How many moles of Na2O are needed to make 1000 g of NaOH?

1000g (mole/40 g) = 25 moles NaOH

25 moles NaOH

c. What would happen if 18 g of water were mixed with 18 g of sodium oxide?

18 g of H2O is 1 mole

18 g of Na2O = 18g(mole/62g) =0.29 moles

but according to the equation water and sodium oxide react in a 1:1, so we have too much water . All 0.29 moles of  Na2O will react with only 0.29 moles of water to create twice as many moles of NaOH (see ratio). Answer = 0.58 moles of NaOH result.

0.58 moles(40 g/mole) = 23.2 g

3.     Balance and answer the questions that follow:

C + 2 H2àCH4

a. How many moles of CH4 can be made from 7.0 g of H2?

7.0 g of H2= 3.5 moles H2. How? Divide by molar mass of diatomic hydrogen.

From the ratio, only half as many moles of methane will be produced so answer = 1.75 moles CH4.

b. What weight of H2 is needed to react with 5.0 g of C?

Convert 5.0 g of C to moles: 0.417 moles C

From ratio: 0.834 moles of H2 is needed.

Convert to mass by multiplying by molar mass:

1.66 g H2

c. What would happen if 20 g of hydrogen were mixed with 20 g of carbon?

20 g of C(mole/12 g) = 1.67 mole C. This amount needs 2(1.75 moles) = 3 moles of H2.(see ratio). We clearly have an excess of H2 since 20 g = 10 moles.

The amount of CH4 produced will depend on C, the limiting reagent, and only 1.67e moles of CH4 will be produced.

1.67 moles (16g/mole) = 27 g

Explanation:

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