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2 years ago

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solmaris [256]2 years ago

8 0

The answer is option D, limiting reactant.

Oftentimes in chemistry, there are leftovers of certain chemicals in reactions. When not all of a certain reactant is used up in a reaction, it is known as the excess reactant (since there is an excess of it). The reactant that's completely used up is known as the limiting reactant (since it limits the reaction from going any further)

-T.B.

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A sample of oxalic acid (a diprotic acid of the formula H2C2O4) is dissolved in enough water to make 1.00 L of solution. A 100.0

OleMash [197]

<u>Answer:</u> The mass of original oxalic acid sample is 6.75 grams

<u>Explanation:</u>

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2C_2O_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=?M\\V_1=100.0mL\\n_2=1\\M_2=0.750M\\V_2=20.0mL$

Putting values in above equation, we get:

$2\times M_1\times 100.0=1\times 0.750\times 20.0\\\\M_1=\frac{1\times 0.750\times 20.0}{2\times 100.0}=0.075M$

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Given mass of oxalic acid = ? g

Molar mass of oxalic acid = 90 g/mol

Molarity of solution = 0.075 M

Volume of solution = 1.00 L

Putting values in above equation, we get:

$0.075M=\frac{\text{Mass of oxalic acid}}{90g/mol\times 1L}\\\\\text{Mass of oxalic acid}=(0.075\times 90\times 1)=6.75g$

Hence, the mass of original oxalic acid sample is 6.75 grams

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2 years ago

HELPPPP PLEASEE!!!!!!!

OlgaM077 [116]

I think the answer is yes

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