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3 years ago

Please help me balance this equation!!!! _KOH + _HBr → -KBr + _H2O

Chemistry

2 answers:

Naddika [18.5K]3 years ago

8 0

Answer: KOH + HBr = H2O + KBr

Explanation: hope this helps

mart [117]3 years ago

6 0

I think it's already balanced so just put a 1 on all of them .

K - 1, 1

O - 1, 1

H - 2, 2

Br - 1, 1

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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.

Answer: 28.0 %

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 %

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