calculate the mass of oxalic acid dihydrate C2H2O4.2H2O required to react with 30 ml of 0.20 M sodium hydroxide solution?

H2C2O4.2H2O + 2NaOH ==> Na2C2O4 + 4H2O

moles NaOH = M x L = 0.20 M x 0.030 L = ??

moles H2C2O4.2H2O required = moles NaOH x 1/2

mass H2C2O4.2H2O = moles x molar mass.

Post your work if you get stuck. Check my work.

Well, if we're going to calculate the mass of oxalic acid dihydrate, we need to break out our trusty periodic table. Let's do some chemistry math with a touch of clownish humor!

First, let's start with the sodium hydroxide (NaOH) solution. Since you mentioned it's 0.20 M, which stands for molarity, we know there are 0.20 moles of NaOH in 1 liter of solution. And since you have 30 ml (or 0.03 liters), we can calculate the number of moles of NaOH using the equation:

moles = concentration (M) x volume (L)

So, moles of NaOH = 0.20 M x 0.03 L = 0.006 moles of NaOH.

Now, let's move on to oxalic acid dihydrate (C2H2O4.2H2O). The molar ratio between oxalic acid and sodium hydroxide is 1:2, based on the balanced chemical equation. This means that 1 mole of oxalic acid dihydrate reacts with 2 moles of sodium hydroxide.

Therefore, if we have 0.006 moles of NaOH, we would need half that number, aka 0.003 moles, of oxalic acid dihydrate.

Now, to find the mass, we need to multiply the moles by the molar mass of oxalic acid dihydrate. The molar masses are as follows:

C: 12 g/mol
H: 1 g/mol
O: 16 g/mol
H2O: 18 g/mol

So, the molar mass of oxalic acid dihydrate is:

(2 x C) + (2 x H) + (4 x O) + (2 x H2O)
(2 x 12) + (2 x 1) + (4 x 16) + (2 x 18)
= 90 g/mol

Finally, multiplying the molar mass by the number of moles, we get:

mass = moles x molar mass
mass = 0.003 moles x 90 g/mol

Drumroll, please! The answer is:

mass = 0.27 grams of oxalic acid dihydrate!

Now, wasn't that a fun and educational trip through chemistry? I hope it put a smile on your face, just like a clown!

To calculate the mass of oxalic acid dihydrate (C2H2O4.2H2O) required to react with 30 ml of 0.20 M sodium hydroxide (NaOH) solution, we need to use the balanced chemical equation between oxalic acid dihydrate and sodium hydroxide.

The balanced equation is as follows:
C2H2O4.2H2O + 2NaOH -> 2H2O + Na2C2O4

From the balanced equation, we can see that one mole of oxalic acid dihydrate reacts with two moles of sodium hydroxide.

First, let's calculate the number of moles of sodium hydroxide in 30 ml of 0.20 M NaOH solution:

Molarity (M) = moles/volume (L)
0.20 M = moles/0.030 L (converted 30 ml to L)
moles = 0.020 moles

From the balanced equation, we know that the ratio between C2H2O4.2H2O and NaOH is 1:2. Therefore, the number of moles of oxalic acid dihydrate required will also be 0.020 moles.

Now, we need to calculate the molar mass of oxalic acid dihydrate:
C = 12.01 g/mol (2 atoms) = 24.02 g/mol
H = 1.01 g/mol (4 atoms) = 4.04 g/mol
O = 16.00 g/mol (4 atoms + 2 atoms) = 96.00 g/mol

Molar mass of oxalic acid dihydrate (C2H2O4.2H2O) = 24.02 g/mol + 4.04 g/mol + 96.00 g/mol = 124.06 g/mol

Finally, we can calculate the mass of oxalic acid dihydrate needed:

Mass (g) = moles x molar mass
Mass = 0.020 moles x 124.06 g/mol ≈ 2.48 g

Therefore, approximately 2.48 grams of oxalic acid dihydrate would be required to react with 30 ml of 0.20 M sodium hydroxide solution.

To calculate the mass of oxalic acid dihydrate (C2H2O4.2H2O) required to react with a given volume and concentration of sodium hydroxide (NaOH) solution, we need to use the concept of stoichiometry.

Step 1: Write the balanced chemical equation for the reaction between oxalic acid dihydrate and sodium hydroxide:

C2H2O4.2H2O + 2NaOH → Na2C2O4 + 4H2O

Step 2: Identify the stoichiometric ratio between oxalic acid dihydrate (C2H2O4.2H2O) and sodium hydroxide (NaOH). From the balanced equation, we can see that for every 1 mole of oxalic acid dihydrate, we need 2 moles of sodium hydroxide.

Step 3: Convert the given volume of sodium hydroxide solution to moles. We are given that the sodium hydroxide solution is 0.20 M (0.20 moles per liter). Since we have 30 ml (0.03 liters) of the solution, the moles of NaOH can be calculated as follows:

Moles of NaOH = volume (in liters) x concentration (in moles per liter)
= 0.03 L x 0.20 M
= 0.006 moles of NaOH

Step 4: Use the stoichiometric ratio to determine the moles of C2H2O4.2H2O required. We know from the balanced equation that for every 2 moles of NaOH, we need 1 mole of C2H2O4.2H2O. Thus, the moles of C2H2O4.2H2O required can be calculated as:

Moles of C2H2O4.2H2O = 0.006 moles of NaOH / 2
= 0.003 moles of C2H2O4.2H2O

Step 5: Calculate the molar mass of C2H2O4.2H2O. The atomic masses of carbon (C), hydrogen (H), and oxygen (O) are 12.01 g/mol, 1.01 g/mol, and 16.00 g/mol, respectively. There are two carbon atoms, two hydrogen atoms, eight oxygen atoms, and two water molecules in one mole of oxalic acid dihydrate. Hence, the molar mass of C2H2O4.2H2O can be calculated as:

Molar mass of C2H2O4.2H2O = (2 x (12.01 g/mol)) + (2 x (1.01 g/mol)) + (8 x (16.00 g/mol)) + (2 x ((2 x (1.01 g/mol)) + 16.00 g/mol))
= 90.03 g/mol

Step 6: Calculate the mass of C2H2O4.2H2O required. The mass can be calculated using the equation:

Mass = Moles x Molar mass
= 0.003 moles x 90.03 g/mol
= 0.2701 grams

Therefore, approximately 0.2701 grams of oxalic acid dihydrate (C2H2O4.2H2O) is required to react with 30 ml of 0.20 M sodium hydroxide solution.