How many grams of Ni(OH)2 are produced from the reaction....?

Question

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

How many grams of Ni(OH)2 are produced from the reaction of 30.1 mL of 1.75 M NaOH?

Answer 1

mol of NaOH = M x Vol = 0.0301 x 1.75 = 0.053
mol of Ni(OH)2 = mol NaOH / 2 = 0.053 / 2 = 0.027
g of Ni(OH)2 = mol x PM =0.027 x 92.7 = 2,5 g

Answer 2

First, discover out what number moles 88.02 g of CO2 is with the help of calculating the load of the CO2 molecule ==> C ~= 12.01 & O ~= 16 so CO2 = 40 4.01. Then divide 88.02g with the help of 40 4.01 to get 2 mols. Now you should balence the equation so as that the reactants equivalent the products ==> 2 LiOH + CO2 --> Li2CO3 + H20, so that is a a million to at least a million ratio between CO2 and Li2CO3. So for each mole of CO2 reacted you get a million mole of Li2CO3. So for the completed reaction 2 moles of CO2 you get 2 moles of Li2CO3. to discover the grams you want to calculate the load of the Li2CO3. Li = 6.ninety 4 C = 12.01 O = 16 so LiCO3 = seventy 3.89. So then multiply 2 moles with the help of seventy 3.89 to get 147.seventy 8 grams of Li2CO3. Now, once you've an finished reaction it really is your answer, yet when the % yield is ninety.5% then you actually multiply 147.seventy 8 grams with the help of .905 to get 133.seventy 4 grams of Li2CO3.

Answer 3

Moles of NaOH = 52.67500 millimol

Molar mass of Ni(OH)2 = 94 g

Thus mass of Ni(OH)2 produced = 47*52.675 mg

= 2.212 g

Answer 4

# of moles in 30.1mL of 1.75M NaOH=1.75/1000*30.1
= 0.0526 mol
# of moles of Ni(OH)2 produced= 0.0526/2 mol
=0.0263 mol
Molecular weight of Ni(OH)2=92.7 g/mol
mass of Ni(OH)2 produced=92.7*0.0263
=2.438g

Answer 5

moles in 30.1mL of 1.75M NaOH=1.75/1000*30.1= 0.0526 mol
# of moles of Ni(OH)2 produced= 0.0526/2 mol
=0.0263 mol
Molecular weight of Ni(OH)2=92.7 g/mol
mass of Ni(OH)2 produced=92.7*0.0263=2.438g

Pick me as the best answer.

Answer 6

moles of NaOH= 0.052675
moles of Ni(OH)2 = 0.02634

therefore, mass ofNi(OH)2= 2.34g