If you are a chemistry genius....please help me :) A precip. reaction stoichiometry problem!!?

Question

I'm having a problem with this question and I have nothing else to check it by.
What mass of Na2CrO4 is required to precipitate all of the silver ions from 75.0 mL of a .100 M solution of AgNO3?
My answer is 1.21g.
Is this correct and if it is not, Why?
THANKS SO MUCH!!!!

Answer 1

Find the quantity of AgNO3 moles:

n = C*V, n = 0.1 * 0.075 = 0.0075 mol

According to the chem. equation:

2AgNO3 + Na2CrO4 --> 2NaNO3 + Ag2CrO4

2 mol AgNO3 react with 1 mol Na2CrO4
0.0075 mol AgNO3 react with 0.0075/2 = 0.00375 mol Na2CrO4

The relative molar mass of Na2CrO4 equals:

Mr = 2*23 + 52 + 4*16 = 162

Now you can find the mass of Na2CrO4:

m = n*Mr = 0.00375 * 162 = 0.6075 g

Answer 2

I got this off of www.Jiskha.com that is a homework help site.

There is a four-step procedure for doing most of these although the process must be tweaked for some problems.
Step 1. Write the balanced equation.
2AgNO3 + Na2CrO4 ==> Ag2CrO4 + 2NaNO3

Step 2. Convert what you have (in this case AgNO3) to mols.
L x M = mols. 0.0750L x 0.100M = 0.00750 mols.

Step 3. Convert mols of what you have (in this case AgNO3) to mols of what you want (in this case Na2CrO4). To do this, use the coefficients in the balanced chemical equation.
0.00750 mols AgNO3 x (1 mol Na2CrO4/2 mols AgNO3) = 0.003750 mols Na2CrO4. Note the numbers come from the coefficients of the equation. They are placed in the numerator and denominator so that the mols of what you want are on top and the mols of what you have cancel (one in the numerator and one in the denominator). That is the fraction used gives the units you want.

Step 4. Now convert mols in step 3(in this case mols Na2CrO4) to grams. mols x molar mass = g; 0.00375 mols Na2CrO4 x molar mass Na2CrO4 = g Na2CrO4.

Answer 3

Atomic weights: Na = 23 Cr = 52 O = 16 Na2CrO4 = 162

Na2CrO4 + 2AgNO3 ===> 2NaNO3 + Ag2CrO4

75.0mLsoln x 0.100molAgNO3/1000mLsoln x 1molNa2CrO4/2molAgNO3 x 162gNa2CrO4/1molNa2CrO4 = (75.0)(0.100)(162)/(1000)(2) = 0.608gNa2CrO4

The first factor is the given amount of solution. The second factor comes from the molarity, 0.100M. The mL of solution cancel, leaving moles of AgNO3. The third factor comes from the balanced equation. The moles of AgNO3 cancel, leaving moles ofNa2CrO4. The fourth factor comes from the formula weight of Na2CrO4. The moles of Na2CrO4 cancel, leaving grams of Na2CrO4.

Answer 4

Na2SiO3(s) + 8 HF(aq) ? H2SiF6(aq) + 2 NaF(aq) + 3 H2O(l) it particularly is the balanced equation and the molar ratios of each and each compound to the others is right here: a million:8:a million:2:3, respectively 0.24moles Na2SiO3 x (8HF / 1Na2SiO3) = a million.92moles HF 0.467moles HF x (2NaF / 8HF) x 42g/mole = 4.9g NaF 0.871g HF / 20g/mole = 0.044moles HF 0.044moles HF x (1Na2SiO3 / 8HF) x 122.1gmole = 0.672g Na2SiO3 you will desire to truly verify to're coping with moles. one won't be able to evaluate mass of compound A with mass compound B by actuality the form of debris may well be diverse. as you with any success be attentive to, 1mole H2 = 2g/mole and 1mole Pb = 207g/mole yet they have an comparable type of atoms stoichiometry is extremely ordinary in case you do now no longer placed the worry of chemistry into it. ex. you're to place vehicles on the same time. 2 windshields + 4tires + 1body = 1car balanced equation in case you have 55windshields what proportion vehicles are you able to're making? fifty 5 windshields x (1car / 2windshields) = 27.5cars in case you have 55windshields and one hundred tires, what proportion vehicles are you able to're making? for each 2 windshields you like 4 tires fifty 5 windshields will require one hundred ten tires we in rapidly forward words have one hundred. consequently the tires will decrease returned the form of vehicles which may well be produced. it particularly is the proscribing reactant you different than for mght with any success be attentive to the thank you to bypass from moles to grams so, bear in strategies, come across or convert to moles, evaluate moles, convert to grams if required.

Answer 5

Lets see...
You have .075 L of a .100 M solution. That means that you have .075*.100=.0075 moles of Silver Nitrate, and thus .0075 moles of Silver in solution. Since the charge of Silver in this case is 1+, and the charge of the Chromate is 2-, the resulting precipitate will be Ag2CrO4. One ion of chromate will take two ions of silver, so you only need .00375 moles of chromate, and thus only .00375 moles of sodium chromate (since there is one ion of chromate per molecule of sodium chromate). The molar mass of sodium chromate is 2(Na)+ (Cr)+ 4(O)=2(22.99)+(51.94)+4(16)=45.98+59.14+64=169.12 g/mol. You need .00375 moles, so that in grams is 169.12x.00375=.634 grams.