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zalisa [80]
3 years ago
8

Please help me #6!!!!!!!!

Chemistry
1 answer:
Vesnalui [34]3 years ago
3 0
I believe that the answer is 12 because there is already 3 O molecules and since its in parentheses with 3 outside it that means that there are 3 of those CO molecules meaning that for every 1 CO there will be 3 O’s so 3, four times Is 12
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2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3H2(g)a. Determine the volume (mL) of 15.0 M sulfuric acid needed to react with 45.0 g of
Ira Lisetskai [31]

Answer:

a. 167 mL.

b. 39.3 %.

Explanation:

Hello!

In this case, for the undergoing chemical reaction, since 45.0 g of aluminum react, based on the 2:3 mole ratio with sulfuric acid, we can compute the required moles  as shown below:

n_{H_2SO_4}=45.0gAl*\frac{1molAl}{27.0gAl} *\frac{3molH_2SO_4}{2molAl} =2.50molH_2SO_4

Next, since the molarity of a solution is computed based on the moles and volume (M=n/V), we can compute the required volume of sulfuric acid as shown below:

V=\frac{n}{M}=\frac{2.50mol}{15.0mol/L}=0.167L

That in mL is 167 mL.

Moreover, for the percent yield, we compute the grams of aluminum sulfate that are produced based on the required 2.50 moles of sulfuric acid:

m_{Al_2(SO_4)_3}=2.50molH_2SO_4*\frac{1molAl_2(SO_4)_3}{3molH_2SO_4}*\frac{342.15gAl_2(SO_4)_3}{1molAl_2(SO_4)_3} \\\\m_{Al_2(SO_4)_3}=285.13gAl_2(SO_4)_3

Therefore the percent yield is:

Y=\frac{112g}{285.13g}*100\%\\\\Y=39.3\%

Best regards!

6 0
2 years ago
Ph of 0.1M solution of acetic acid will be hogher than that of 0.1MHcl​
Aliun [14]

pH=-lg[H⁺]

[H⁺] is the molar concentration of hydrogen protons

both solutions have the same concentration of the monoprotic acid so they will both have the same pH

pH=-lg0.1=1 because 10 at the power of -1 is equal to 0.1 but the sign is changed

8 0
2 years ago
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