3 years ago

Look at the chemical equation:

Chemistry

1 answer:

zhuklara [117]3 years ago

4 0

Answer:

The correct answer is 6 i believe

Explanation:

6 would be on both sides on H while balancing the equation 2 would be for N

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What volume in milliliters of 1.420 M sulfuric acid is needed to neutralize 3.209 g of aluminum hydroxide 3 H 2 SO 4 (aq)+2 Al(O

nadya68 [22]

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$V=43.46mL$

Explanation:

Hello!

In this case, since the reaction between sulfuric acid and aluminum hydroxide is:

$3H_2SO_4+2Al(OH)_3\rightarrow Al_2(SO_4)_3+6H_2O$

Whereas the ratio of sulfuric acid to aluminum hydroxide is 3:2; thus, we first compute the moles of sulfuric acid that complete react with 3.209 g of aluminum hydroxide:

$n_{H_2SO_4}=3.209gAl(OH)_3*\frac{1molAl(OH)_3}{78.00gAl(OH)_3} *\frac{3molH_2SO_4}{2molAl(OH)_3} \\\\n_{H_2SO_4}=0.0617molH_2SO_4$

Then, given the molarity, it is possible to obtain the milliliters as follows:

$V=\frac{n}{M}=\frac{0.0617mol}{1.420mol/L}*\frac{1000mL}{1L}\\\\V=43.46mL$

Best regards!

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