I've stuck with these two stoich questions
A mixture is prepared by adding 20 ml of 0.200 M Na3PO4 to 30 ml of 0.150 Ca(NO3)2. What weight of calcium phospate will be formed?
and last question would be, How many milliliters of 0.30 M NiCl2 solution are needed to react completely with 25 ml of 0.10 M Na2CO3 solution? How many grams of NICO3 will be formed?
Molarity = number of moles / Liters
The first question is a limiting reagent problem. You work it the same way as the previous LR problem.
The second one is not a LR problem but a simple stoichiometry problem. So instead of converting two reactants to mols of product and choosing which one to use you have only the one reagent to worry with and there is no choice.
The idea with this homework help site is that we show you how to do one and you use that as a template to solve the other problems you have.
I shall be happy to explain further if you tell be exactly what you don't understand. You have the information to solve these two problems.
To solve these stoichiometry problems, you need to use the concept of mole ratios and the formula for calculating the number of moles using molarity.
For the first question, let's determine the balanced equation for the reaction between Na3PO4 and Ca(NO3)2:
2 Na3PO4 + 3 Ca(NO3)2 -> Ca3(PO4)2 + 6 NaNO3
Now, let's calculate the number of moles of each reactant:
Number of moles of Na3PO4 = Molarity x Volume = 0.200 M x 0.020 L = 0.004 moles
Number of moles of Ca(NO3)2 = 0.150 M x 0.030 L = 0.0045 moles
Since the stoichiometric ratio between Na3PO4 and Ca3(PO4)2 is 2:1, we can conclude that the limiting reactant is Na3PO4 because it produces fewer moles of the desired compound.
To determine the weight of calcium phosphate formed, you need to know the molar mass of Ca3(PO4)2, which is approximately 310 g/mol. Now, using the stoichiometric ratio, we can calculate the weight of calcium phosphate:
Weight = Number of moles x Molar mass = 0.004 moles x 310 g/mol = 1.24 grams
For the second question, let's write the balanced equation for the reaction between NiCl2 and Na2CO3:
NiCl2 + Na2CO3 -> NiCO3 + 2 NaCl
Using the same approach, let's calculate the number of moles of each reactant:
Number of moles of NiCl2 = Molarity x Volume = 0.30 M x (unknown volume in liters)
Number of moles of Na2CO3 = 0.10 M x 0.025 L = 0.0025 moles
Since the stoichiometric ratio between NiCl2 and NiCO3 is 1:1, we can say that the number of moles of NiCl2 in the unknown volume is 0.0025 moles.
To find the volume of NiCl2, we can rearrange the molarity formula:
Molarity = Number of moles / Volume
0.30 M = 0.0025 moles / Volume
Volume = 0.0025 moles / 0.30 M = 0.0083 liters
Since 1 liter is equal to 1000 milliliters, the volume of NiCl2 solution required is 8.3 mL.
Now, let's calculate the mass of NiCO3 formed. First, we need to know the molar mass of NiCO3, which is approximately 117 g/mol. Using the stoichiometric ratio, we can calculate the mass of NiCO3:
Mass = Number of moles x Molar mass = 0.0025 moles x 117 g/mol = 0.29 grams
So, in the second question, 8.3 mL of 0.30 M NiCl2 solution is needed to react completely with 25 mL of 0.10 M Na2CO3 solution, and 0.29 grams of NiCO3 will be formed.