How many grams of Ni(OH)2 are produced from the reaction of 25.0mL of a 0.600M NaOH solution? NiCl2(aq)+2NaOH(aq)→

Ni(OH)2(s)+2NaCl(aq)

NiCl2(aq) + 2NaOH(aq) → Ni(OH)2(s) + 2NaCl(aq)

First get the number of moles of NaOH. Recall that molarity is moles of solute per liter of solution:
M = n / V
Getting n,
n = MV
n = (0.6)(0.025)
n = 0.015 mol NaOH
To get the mol Ni(OH)2, we multiply the mol of NaOH to the ratio found from the reaction. From the reaction, for every 1 mol of Ni(OH)2 produced, 2 mol NaOH are consumed.
0.015 mol NaOH * 1 mol Ni(OH)2 / 2 mol NaOH = 0.0075 mol Ni(OH)2
Finally, to get the mass of Ni(OH)2, multiply 0.0075 mol by the molar mass of Ni(OH)2.

Hope this helps :3

mols NaOH = M x L = ?

Use the coefficients in the balanced equation to convert mols NaOH to mols Ni(OH)2
Now convert mols Ni(OH)2 to g. g = mols x molar mass.

1.25 g Ni(OH)2

Well, the reaction seems like a party between NiCl2 and NaOH, where they come together for a dance and produce Ni(OH)2 as a product, along with 2NaCl. Quite a lively gathering, I must say!

To find out how many grams of Ni(OH)2 are produced, we need to do a little stoichiometry dance of our own. First, let's convert the volume of the NaOH solution to moles by using the molarity:

25.0 mL × 0.600 mol/L = 15.0 mmol

Now, we can use the balanced equation to compare the ratio of moles of NaOH to moles of Ni(OH)2. From the equation, we see that it's a 1:1 ratio, so we have 15.0 mmol of Ni(OH)2.

To find the gram equivalent of that, we need to know the molar mass of Ni(OH)2. Nickel (Ni) has an atomic weight of approximately 58.69 g/mol, while oxygen (O) has a molar mass of around 16.00 g/mol, and hydrogen (H) has a molar mass of about 1.01 g/mol.

Adding all these up, we get:

(58.69 g/mol) + 2 × [(1.01 g/mol) + (16.00 g/mol)] = 92.69 g/mol

We can now calculate the mass of Ni(OH)2 produced:

15.0 mmol × (92.69 g/mol) = 1390 g

So, approximately 1390 grams of Ni(OH)2 is produced from the reaction. That's quite a weighty party favor!

To find out how many grams of Ni(OH)2 are produced from the reaction, you need to use stoichiometry. Here's how you can do it step-by-step:

Step 1: Write the balanced equation for the reaction:
NiCl2(aq) + 2NaOH(aq) → Ni(OH)2(s) + 2NaCl(aq)

Step 2: Determine the moles of NaOH:
To do this, use the formula:
moles = concentration (in M) × volume (in L)
First, convert the given volume of NaOH solution from mL to L:
25.0 mL = 25.0 ÷ 1000 L = 0.025 L
Now, calculate the number of moles of NaOH:
moles NaOH = 0.600 M × 0.025 L = 0.015 mol NaOH

Step 3: Use the stoichiometry of the balanced equation to find the moles of Ni(OH)2 produced:
From the balanced equation, you can see that the reaction produces a 1:1 ratio of Ni(OH)2 to NaOH. Therefore, the moles of Ni(OH)2 produced will also be 0.015 mol.

Step 4: Calculate the molar mass of Ni(OH)2:
The molar mass of Ni is 58.69 g/mol, and for OH it is 17.01 g/mol. Since Ni(OH)2 has 2 OH groups, the molar mass of Ni(OH)2 is:
molar mass Ni(OH)2 = 58.69 g/mol + (2 × 17.01 g/mol) = 92.71 g/mol

Step 5: Calculate the mass of Ni(OH)2 produced:
Use the formula:
mass = moles × molar mass
mass Ni(OH)2 = 0.015 mol × 92.71 g/mol ≈ 1.39 g

Therefore, approximately 1.39 grams of Ni(OH)2 are produced from the reaction of 25.0 mL of a 0.600 M NaOH solution.