I need a little help with stoichiometric problems?

Question

1. What is the mass of O in 148 g of Calcium Hydroxide (Ca(OH)2)?
2. 2Na(s) + 2H2O(l) ----> 2NaOH(aq) + H2(g)
Elemental sodium reacts with water to form hydrogen gas as shown above. If a sample of sodium reacts completely to form 20 liters of hydrogen gas, measured at standard temperature and pressure, what was the mass of the sodium?
3. CaCO3(s) + 2H -----> Ca^+2(aq) + H2O(l) + CO2(g)
If the reaction above took place at standard temperature and pressure and 150 grams of CaCO3(s) were consumed, what was the volume of CO2(g) produced ad STP?
4. A gaseous mixture at 25 degrees celcius contained 1 mole of CH4 and 2 moles of O2 and the pressure was measured at 2 atm. The gases then underwent the reaction shon below.
CH4(g) + 2O2(g) -----> CO2(g) + 2H2O
What was the pressure of the container after the reaction had gone into completion and the temperature was allowed to return to 25 degrees celcius?

Please explain to me how to solve them. And if you're some that'll answer "Do your own homework", don't bother answering.

Please, someone answer my problems. I need help. I'll give the first person to answer my question 10 pts.

Answer 1

1.) To find the mass of 1 element in a sample of a molecule:

Step 1: Find the molar mass of the molecule.
Ca(OH)2 has 1 atom of Ca (40.078), 2 molecules of O (16), 2 molecules of H (2) for a molar mass of 74.078.

Step 2. Given the mass of the sample, and the molar mass, find the number of moles of the sample. From that, find the # of moles of O, then multiply by the molar mass of O (16).
# of moles = mass of sample / molar mass
# of moles = 148 / 74.078 = 1.99789 (we'll just round to 2).

If there are 2 moles of sample, and 2 moles of Oxygen per mole of sample, then there are 4 moles of Oxygen. 4 moles of Oxygen * 16 g/mol = 64g of Oxygen in the sample (solution!)

2.) Step 1: Use the ideal gas law to find the number of moles of Hydrogen.
PV = nRT.
where P = pressure = 100 kPa
R = ideal gas constant = 8.314472 L · kPa · K-1 · mol-1
V = volume = 20 L
T = Temperature = 0 °C = 273.15 K
n = number of moles.
n = PV / RT = (100 * 20) / (8.314472 * 273.15)
n = 0.8806 moles of H2

Givem the # of moles of H2, find the # of moles of sodium.

Given your equation, there is 2 moles of Na for 1 mole of H2. 2 * 0.8806 = 1.7612 moles of Na.

3 is the same as 2, but reversed = you convert the mass of CaCO3 to moles, then use the # of moles and STP to find the volume.

4 looks to be a trick question. If the volume doesn't change, the begin and end temperature are the same, and the # of moles are the same on each end of the reaction, then pressure must be the same.
Proof:
P1V1 = n1RT1 and P2V2 = n2RT2

Rewrite for R:
R = P1V1/n1T1 = P2V2/n2T2

Let V1 = V2, n1 = n2 = 3, and T1 = T2 = 25 C:
P1V1/n1T1 = P2V2/n2T2
P1V1/n1T1 = P2V1/n1T1
V1, n1, and T1 cancel...
P1 = P2.

Answer 2

hi There, Question1: one million. enable's discover the MW of NH3, 14g/mol +3g/mol = 17 g/mol 2. discover the variety of moles of NH3= 22.7 g * mol/17g = one million.34 mol of NH3 3. If one million.34 mol of NH3 is what we've, then we would choose one million.34/2 mol (0.sixty seven mol) of H2SO4. 4. discover MW of H2SO4 = 2g/mol +32g/mol + 16g/mol *4 = ninety 8 g/mol 5. how many moles of H2SO4 can we've from fifty 4.8 g? fifty 4.8g *mol/98g = 0.559 mol of H2SO4 is accessible. properly, the quantity of H2SO4 will surely be under 0.sixty seven mol that's under the mandatory variety of moles to react each and all the NH3 accessible. What does this recommend? It ability that H2SO4 is the restricting reagent. hence, you may in basic terms make the reaction bypass as some distance because of the fact the quantity of accessible H2SO4 facilitates you. 3. in view that we've 0.559 mol of H2SO4, we are able to apply 0.559 *2 moles ( one million.118 moles) of NH3. 4. Now the product: the ration of H2SO4 to (NH4)2SO4 is one million:one million hence, with 0.559 mol of H2SO4, 0.559 mol of (NH4)2SO4 is made 5. how many grams is that? MW of (NH4)2SO4 is : [(14 +4) *2 + 32 +sixteen*4] g/mol = 132 g/mol Grams of (NH4)2SO4 is " 0.559 mol * 132 g/mol = seventy 3.8 g of (NH4)2SO4 is made. This became a restricting reagent concern! question 2: one million. assume mass is 100g then the probability is the comparable because of the fact the lots: fifty 4.53g * mol/12g C: 9.15 g * mol/g H : 36.32 g *mol/sixteen g O 2. The moles are 4.54mol of C : 9.15mol of H and 2.27mol of O 2a: looking the ration by dividing with the backside: C2H4O 3. Giving us C2H4O 4. MW of C2H4O is: 44g/mol so two times it extremely is 88g/mol for this reason, MW of 88amu is C4H8O2 question 3 A is actual B) one million. if 6.5 moles of NO2 is produced then 6.5 *4 moles of HNO3 became reacted because of the fact ration HNO3 to NO2 is 4:one million so which you quite often could look on the coefficient and relate the moles of the two compounds. you may write it as stick to: 6.5 moles NO2 * 4 moles of HNO3 / one million mole of NO2 2. Now discover the MW of HNO3 : [one million+ 14 + 3*sixteen] g/moles = sixty 3 g/mole 3. discover grams of HNO3 : 6.5 moles * 63g/mole = 409.5 g of HNO3 became used because of the fact the reactant! i'm hoping all of it is clever!