How many grams in 4.5 moles of H2O?

How many atoms are in 64 g sulfur (S)?

Katen [24]

Answer:

there are 6.022*1023 atoms

Explanation:

32 g S * (1 mole S/32 g S) * (6.022*1023 atoms/1 mole S)= 6.022*1023 atoms

6 0

3 years ago

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The reaction of ethane gas (C2H6) with chlorine gas produces C2H5Cl as its main product (along with HCl). In addition, the react

Kazeer [188]

Answer:

The percent yield of chloro-ethane in the reaction is 82.98%.

Explanation:

$C_2H_6+Cl_2\rightarrow C_2H_5Cl+HCl$

Moles of ethane = $\frac{300.0 g}{30 g/mol}=10 mol$

Moles of chlorine gases = $\frac{650.0 g}{71 .0 g/mol}=9.1549 mol$

As we can see that 1 mol of ethane react with 1 mole of chlorine gas.the 10 moles will require 10 mole of chlorine gas, but only 9.1549 moles of chlorine gas is present.

This means that chlorine gas is in limiting amount and amount of formation of chloro-ethane will depend upon amount of chlorine gas.

According to reaction , 1 mol of chloro ethane gives 1 mol of chloro-ethane.

Then 9.1549 moles of chlorien gas will give:

$\frac{1}{1}\times 9.1549 mol=9.1549 mol$ of chloro-ethane

Mass of 9.1549 moles of chloro-ethane:

9.1549 mol × 64.5 g/mol = 590.4910 g

Theoretical yield of chloro-ethane: 590.4910 g

Given experimental yield of chloro-ethane: 490.0 g

$\% Yield=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$\%Yield (C_2H_5Cl)=\frac{490.0 g}{590.4910 g}\times 100=82.98\%$

The percent yield of chloro-ethane in the reaction is 82.98%.

6 0

3 years ago

When determining the specific heat from a given metal in a calorimeter the mass of the water remains constant, but the mass of t

alexandr1967 [171]

The heat is hotter than cold

6 0

3 years ago

What is the mass of potassium chloride when 6.75 g of potassium reacts with an excess of chlorine gas? the balanced chemical equ

lorasvet [3.4K]

The balanced equation for the above reaction is;
2K + Cl₂ ---> 2KCl
Stoichiomtery of K to KCl is 2:2
Potassium is the limiting reactant which is fully consumed in the reaction. The amount of product formed depends on amount of limits reactant present.
Number of moles of K reacted - 6.75 g/ 39 g/mol = 0.17 mol
Therefore number of KCl moles formed - 0.17 mol
Mass of KCl formed - 0.17 mol x 74.5 g/mol = 12.67 g

6 0

3 years ago

How many grams of Cl2 are in 0.345L at STP?

blsea [12.9K]

Answer: 1.09 g

Explanation:

If we use the approximation that 1 mole is 22.4 L, then setting up a proportion,

1/22.4 = x/0.345 (x is the number of moles in the sample)
x = 0.0154 mol

Since the mass of a mole of chlroine is about 70.9 g/mol, (0.0154)(70.9) = 1.09 g (to 3 s.f.)

3 0

2 years ago