For the following equation to be bal

A sample of sodium reacts completely with 0.497 kg of chlorine, forming 819 g of sodium chloride. what mass of sodium reacted?

Slav-nsk [51]

Sodium reacts to chlorine and gives NaCl. The balanced reaction is given below:

2Na + Cl₂→ 2NaCl. Two moles Na reacts with one mole Cl₂ and produces two moles of NaCl. Atomic mass of Na= 23, Molar mass of Cl₂= 71, molar mass of NaCl=58.5.

So, 46 g Na reacts with 71 g of Cl₂ and produces (2 X 58.5)g = 117 g of NaCl. As per question Na reacts completely which means Na is the limiting reagent. So, number of moles of Na reacts = number moles of NaCl produced.

NaCl produced= 819 g= (819/58.5) moles= 15.69 moles. Therefore, 15.69 moles = 15.69 X 23 g=360.87 g of Na reacted.

8 0

3 years ago

The magnetic field of an object exerts a force on any magnetic object around it.

nika2105 [10]

I believe the answer is true

3 0

2 years ago

Read 2 more answers

Ammonia (NH3) is one of the most common chemicals produced in the united states. it is used to make fertilizer and other product

Greeley [361]

Answer:

a) N2 is the limiting reactant

b) 1215.9 grams NH3

c) 2283.6 grams H2

Explanation:

Step 1: Data given

Mass of N2 = 1000 grams

Molar mass N2 = 28 g/mol

Mass of H2 = 2500 grams

Molar mass H2 = 2.02 g/mol

Step 2: The balanced equation

N2(g) +3H2(g) → 2NH3(g)

Step 3: Calculate moles N2

Moles N2 = mass N2 / molar mass N2

Moles N2 = 1000 grams / 28.0 g/mol

Moles N2 = 35.7 moles

Step 4: Calculate moles H2

Moles H2 = 2500 grams / 2.02 g/mol

Moles H2 = 1237.6 moles

Step 5: Calculate limiting reactant

For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3

N2 is the limiting reactant. There will be consumed 35.7 moles.

H2 is in excess. There will react 3*35.7 = 107.1 moles

There will remain 1237.6 - 107.1= 1130.5 moles H2

This is 1130.5 * 2.2 = 2283.6 grams H2

Step 6: Calculate moles NH3

For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3

For 35.7 moles N2 we'll have 2*35.7 = 71.4 moles NH3

Step 7: Calculate mass NH3

Mass NH3 = moles NH3 * molar mass NH3

Mass NH3 = 71.4 moles * 17.03 g/mol

Mass NH3 = 1215.9 grams NH3

7 0

3 years ago

4. The mass of a silver liberty (coin) was measured three times and each measurement was made

Vaselesa [24]

Answer:

The average mass of the gold coin reported to five significant figures, even though you had to divide by "3" to obtain the average.

Explanation:

Given that,

The mass of a silver liberty was measured three times and each measurement was made to five digits.

The mass values are,

$m_{1}=12.519\ g$

$m_{2}=12.521\ g$

$m_{3}=12.496\ g$

The average mass of the gold coin is 12.512 g.

We know that,

Significant figures :

Significant figures is explained the nonzero, leading zero and zeros between two nonzero digits.

For example,

The digits is 0.003405

In the digit, 345 = nonzero

0.00 = leading zero

The average mass of the gold coin reported to five significant figures because average mass is 12,512. All digits are nonzero.

Nonzero digit are significant.

Hence, The average mass of the gold coin reported to five significant figures, even though you had to divide by "3" to obtain the average.

7 0

3 years ago

A sample of neon gas has a volume of 752 mL at 298 Kelvin (K). What will be the volume at 373 K if

Firdavs [7]

Answer:

V2 = 0.941 L

Explanation:

This follows the Charles' law, a gas law.

Using this eq $\frac{V1}{T1} =\frac{V2}{T2}$

From the given, the missing variable is V2

$\frac{752 ml}{298 K} =\frac{V2}{373 K}$ , change ml to SI unit of Volume (L)

752 ml = 0.752 L

$\frac{752 L}{298 K} =\frac{V2}{373 K}$ , then cross multiply

$280.496 L (K) =298K (V2)$

$\frac{280.496 L (K) }{298K} = (V2)$

$V2 = 0.941 L$

8 0

2 years ago