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user100 [1]
2 years ago
5

Problem Page Gaseous butane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 5.2 g o

f butane is mixed with 32.6 g of oxygen. Calculate the minimum mass of butane that could be left over by the chemical reaction. Round your answer to significant digits.
Chemistry
1 answer:
stich3 [128]2 years ago
7 0

Answer: 0.0 grams

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

a) moles of butane

\text{Number of moles}=\frac{5.2g}{58.12g/mol}=0.09moles

b) moles of oxygen

\text{Number of moles}=\frac{32.6g}{32g/mol}=1.02moles

2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O

According to stoichiometry :

2 moles of butane require 13 moles of O_2

Thus 0.09 moles of butane will require =\frac{13}{2}\times 0.09=0.585moles  of O_2

Butane is the limiting reagent as it limits the formation of product and oxygen is present in excess as (1.02-0.585)=0.435 moles will be left.

Thus all the butane will be consumed and 0.0 grams of butane will be left.

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How many kilograms are in 6.983 moles of baking soda (NaCHO3)?
nasty-shy [4]
It should be about 0.586620881 kilograms
6 0
3 years ago
What's the charge on Ni in the compound NiCi3? *
ss7ja [257]

Answer:

+3

Explanation:

Chlorine is anion with a -1 charge. But they are three chlorine atoms.

-1 * 3 = -3

So they have a -3 charge.

So to balance the compound, the nickel has to be a cation with a +3 charge.

-3 + 3 = 0

Furthermore, a chemical bond always has a 0 charge. Remember that.

Hope it helped! Rate my answer a 5 star if correct.

8 0
2 years ago
Calculate the density of the four solutions. All of the solutions has the volume equal to 25 ml. Red solution has 25.0 g of mass
Scilla [17]

Answer:

              B. Green solution density is 1.06 g/ml and blue solution density is 1.20 g/ml

Explanation:

Density is given as,

                               D = Mass / Volume

Red Solution,

                               D = 25 g / 25 mL

                               D = 1 g/mL

Green Solution,

                               D = 26.5 g / 25 mL

                               D = 1.06 g/mL

Yellow Solution,

                               D = 28.2 g / 25 mL

                               D = 1.128 g/mL

Blue Solution,

                               D = 30 g / 25 mL

                               D = 1.20 g/mL

3 0
2 years ago
What concentrations of acetic acid (pKa = 4.76) and acetate would be required to prepare a 0.15 M buffer solution at pH 5.0? Not
Leni [432]

Answer:

Acetic acid 0,055M and acetate 0,095M.

Explanation:

It is possible to prepare a 0,15M buffer of acetic acid/acetate at pH 5,0 using Henderson-Hasselblach formula, thus:

pH = pka + log₁₀ [A⁻]/[HA] <em>-Where A⁻ is acetate ion and HA is acetic acid-</em>

Replacing:

5,0 = 4,76 + log₁₀ [A⁻]/[HA]

<em>1,7378 =  [A⁻]/[HA] </em><em>(1)</em>

As concentration of buffer is 0,15M, it is possible to write:

<em>[A⁻] + [HA] = 0,15M </em><em>(2)</em>

Replacing (1) in (2):

1,7378[HA] + [HA] = 0,15M

2,7378[HA] = 0,15M

[HA] = 0,055M

Thus, [A⁻] = 0,095M

That means you need <em>acetic acid 0,055M</em> and <em>acetate 0,095M</em> to obtain the buffer you need.

i hope it helps!

7 0
3 years ago
Jane is sliding down a slide. What kind of motion is she demonstrating?
svp [43]
It is translational motion. i know because i found it on this site. its verified too, have a nice day
5 0
3 years ago
Read 2 more answers
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