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With How Much 2.0 Mol/dm3 Hcl Do 0.4g Of Caco3 React With

What volume of Co2 gas is produced at STP when 50 gm of CaCo3 reacts with excess of diluted HCl?

You need to start off by writing out a balanced chemical equation which will show you the relationship between CaCO3 and CO2:CaCO3 + 2HCl → CaCl2 + CO2 + H2OThis equation tells you that for every mol of CaCO3 consumed you will be producing 1 mol of CO2. In other words there is a 1:1 relationship between CaCO3 and CO2.The next step is to determine the amount of CaCO3 you have in moles, and since HCl is in excess, you do not need to worry about which one is the limiting reagent (CaCO3 is limiting since HCl is in excess). You can do that by using the molar mass of CaCO3, which is 100 g/mol:mol CaCO3 = 50 g CaCO3 x 1 mol CaCO3/100 g CaCO3 = 0.5 mol CaCO3We now can calculate the amount in moles of the produced CO2 by using the relationship between them from the balanced chemical equation:0.5 mol CaCO3 x 1 mol CO2/1 mol CaCO3 = 0.5 mol CO2.It might seem redundant doing this as it is a 1:1 relationship for this specific reaction, however it is a good habit to get into even with the simplest of problems.Since CO2 is a gas, we can use the ideal gas equation (PV=nRT) to interconvert between pressure (P), volume (V), mol (n) and temperature (T). The question states that we are under STP conditions which means that the pressure is 1 atm and the temperature is 273.15 K. All the variables other than volume are known:P x V = n x R x T1 atm x V = 0.5 mol x 0.08206 (L.atm/mol.K) x 273.15V = 11.2 LYou could have also solved the problem directly after you calculated the number of moles of CO2 you produced, as under STP conditions, 1 mol of an ideal gas occupies 22.4 L:0.5 mol * 22.4 L/mol = 11.2 L.

In order to completely neutralize 20 mL of a solution of HCl 0.1 M, 40 mL of a solution of NaOH must be added. What is the M of the NaOH solution?

No, this working is incorrect. You tripped up on two points.20 milliliters is what decimal fraction of a liter?milli- means one thousandth, so 20 mL = 20/1000 liters20/1000 = 2/100 = 1/50 literDivide one by fifty:50 ) 1 wont go , result 0.? so try50) 10 won’t go result 0.0? so try50)100 goes twice result 0.02So 0.02 L = 20mL Result for VaYou made the same mistake with Vb = 40 mL SHOULD BE = 0.04 LThe second mistake was more important because in this equation the first two scaling mistakes canceled out!You wrote: 1 mol x 0.1 M x 0.20 L / 1 mol x 0.40 L = 8x10-3M1 x 0.1 x 0.2 / 1 x 0.4 = 0.05M NOT 8 millimol !You multiplied (x) instead of dividing ( / )

With how much 2.0 mol/dm3 HCl do 0.4g of CaCO3 react with?

CaCO3(s) + 2HCl(aq) --> CaCl2(aq) + CO2(g) + H2O(l)
0.400g ........?L

Set up the solution to the problem using the unit-factor method and the appropriate conversion factors.

0.400g CaCO3 x (1 mol CaCO3 / 100.1 g CaCO3) x (2 mol HCl / 1 mol CaCO3) x (1L / 2.0 mol HCl) = 0.0040L HCl ..... or ..... 4.0 mL..... or ..... 4.0 cm³.... . to two significant digits

What mass of NaOH is required to react exactly with 25.0 mL of 1.2 M H2SO4?

This looks like homework! These steps will solve ANY stoichiometric problem:Write the equation for the reaction and balance it. In this case the equation is: 2NaOH + H2SO4 → Na2SO4 + 2H2OConvert the given amount to moles. Molarity, “M” is moles per liter. The given amount is 25 ml of 1.2 M H2SO4. Since Molarity uses Liters, the volume must be converted from ml to L.Use the mole ratio in this case 2 moles of NaOH to ! mole of H2SO4Convert the moles to the required units. In this case the required units are grams. The formula weight in grams er mole.25 ml H2SO4 * 1L/1000 ml * 1.2 Moles/L * 2 moles NaOH/1 mole H2SO4 * 40 g NaOH/ 1 Mole NaOHPerform these calculations and you have the answer! Timothy, When doing homework, the answer is NOT the important thing, the METHOD is! These 4 steps, combined with required changes in units will sove ANY stoichiometric problem easily.

If 5.15 g Fe(NO3) 3 is dissolved in water to make exactly 150.0 ml of solution, what is the molar concentration of the nitrate ions?

Well, first start with the definition of molarity, which is moles per litre, or mol/L. So take the grams of Fe(NO3)3, divide by the molar mass to get moles of Fe(NO3)3. Take that and divide that by the volume in litres and that gives you the molarity of your solution. Then find the ratio of moles of Fe3+ to NO3- (it's 1:3), and multiply that by the molarity to get the molarity of Nitrate in the solution.

A mixture of gases consisting of 7.5 g of CO2, and 10 g of SO2, is in a 2.5 L container at 25 C. What is the total pressure inside the container?

Assuming the mixture to behave as an ideal gas.molecular weight of CO2 = 12+16+16 = 44molecular weight of SO2 = 32+16+16 = 64moles of CO2 = 7.5/44 = 0.17045moles of SO2 = 10/64 = 0.15625n= moles of CO2 + moles of SO2 = 0.15625 + 0.17045 = 0.3267V= 2.5L = 2.5*10^-3  m^3T= 25C = 298KPV = nRTP = nRT/V = 0.3267*8.314*298/(2.5*10^-3) = 323769.10896 Pa    = 3.1953atm

What is the molarity of the solution 5 g NaOH dissolved in 200 ml of water?

The way this question is written makes the answers given so far incorrect. The other answers are most likely what the question is looking for, but not technically correct. To correctly answer this question requires that you know the density of a 2.5% solution of NaOH. Why? Molarity is Moles Solute/Volume of solution. We are given volume of solvent, not of solution. The solution is 2.50% NaOH (5/(200*0.9982)*100%)=2.5047%~2.50%The density of a 2.5% solution of NaOH is 1.0263 g/mLThe total mass of the solution is 204.6 g (5 g NaOH+ 200mL*0.9982 g/mL) so the total volume is 204.6 g/( 1.0263g/mL)=199.36mL=0.19936LNow the Molarity can be calculated.5 g/(39.997 g/mol)/0.19936L=0.62705M~0.627MWe can see that the difference between the molality calculation (which is what the other two answer gave and the molarity calculation only appears in the third significant digit. This difference becomes smaller for more dilute solutions and larger for more concentrated;dr This question is poorly written and requires the student to make assumptions or look up additional data. Unambiguous phrasing of the problem would be something like:5 g of NaOH is mixed with enough water to make a 200 mL solution. What is the molarity of the solution?or5 g of NaOH is mixed with 200 g of water. What is the molality of the solution?

Can you calculate the molarity, molality, and normality of a H2SO4 solution, which is 98% w/w and has a density of 1.8g/mL?

There is a straight formula for converting w/w% into molarity.Molarity={(w/w%)×10×(density in g/ml)}÷(Molar mass of solute in g)Putting values, M=(98×10×1.8)/(98)=18MNow, Normality=Molarity×n-factorPutting values, N=18×2=36N (n-factor is 2 because H2SO4 is a dibasic acid)Molality is a bit tricky because the formula involves tedious calculation.Molality=(1000×Molarity)/{(1000×density in g/ml)-(Molar mass of solute in g×Molarity)}Putting values,m=(1000×18)/{(1000×1.8)-(98×18)}=18000/(1800–1764)=18000/36=500m

If 1.12 g of KOH are dissolved in 250 mL of H2O, what is the pH?

.Convert the units of water first to liters as this gives us an easy gauge of molarity. To get 250ml to L we must multiply by 4. Anything done to one reagent must be done to the other as well to maintain relative concentration so we multiply 1.12g x 4 to get 4.48 g.4.48 grams of KOH contains .08 mol of KOH (from the molar mass/formula weight, The molar mass of KOH is 56.10564 ± 0.00047 g/mol).As KOH is a strong base the [OH-] concentration will equal that of the initial substance due to complete dissociation.pOH = - log [OH-] pOH=-log(.08) pOH=1.096pH+pOH=14pH = 14 - pOH pH = 14 - 1.096pH=12.90