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umka21 [38]
3 years ago
6

A rock weighing 26.0g is placed in a graduated cylinder displacing the volume from 13.2mL to 23.5mL what is the density in grams

/cubic centimeter
Chemistry
1 answer:
IgorC [24]3 years ago
8 0
M=26.0 g
v₁=13.2 mL
v₂=23.5 mL

p=m/(v₂-v₁)

P=26.0/(23.5-13.2)=2.524 g/cm³
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Answer:

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Gaseous butane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water . If of water is produced from
krek1111 [17]

The given question is incomplete. The complete question is :

Gaseous butane reacts with gaseous oxygen gas  to produce gaseous carbon dioxide and gaseous water . If 1.31g of water is produced from the reaction of 4.65g of butane and 10.8g of oxygen gas, calculate the percent yield of water. Be sure your answer has the correct number of significant digits in it.

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Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of butane}=\frac{4.65g}{58g/mol}=0.080moles

\text{Moles of oxygen}=\frac{10.8g}{32g/mol}=0.34moles

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

According to stoichiometry :

13 moles of O_2 require 2 moles of butane

Thus 0.34 moles of O_2 will require=\frac{2}{13}\times 0.34=0.052moles  of butane

Thus O_2 is the limiting reagent as it limits the formation of product and butane is the excess reagent.

As 13 moles of O_2 give = 10 moles of H_2O

Thus 0.34 moles of O_2 give =\frac{10}{13}\times 0.34=0.26moles  of H_2O

Mass of H_2O=moles\times {\text {Molar mass}}=0.26moles\times 18g/mol=4.68g

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

{\text {percentage yield}}=\frac{1.31g}{4.68g}\times 100\%=28.0\%

The percent yield of water is 28.0 %

6 0
4 years ago
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