5. Concentration of 0.24L KMnO4 solution required to oxidize 2.7g of Oxalic Acid??? 5H2C2O4 (aq) + 6H+ + 2MnO4- (aq) --> 10CO2 (g) + 2 Mn2+(aq) +8H20 (l)
4.How many grams of calcium chloride can be prepared if a student uses 500g if calcium carbonate and 2.00L of a 4.00 Mol/L sodium chloride solution in the reaction.??
CaCO3 + 2HCl -> CaCl2 + H2O
8. 2(NH4)3PO4-->3N2+O2+12H2+2PO3
a)name products
b)Name the reactant
c)type of reaction
5. Concentration of 0.24L KMnO4 solution required to oxidize 2.7g of Oxalic Acid??? 5H2C2O4 (aq) + 6H+ + 2MnO4- (aq) --> 10CO2 (g) + 2 Mn2+(aq) +8H20 (l)
Molecular weight:
H2C2O4 = 2+24+64 = 90
2.7 g of oxalic acid is equivalent to
2.7/90=0.03 mole
Number of moles of KMnO4 required
=0.03*2/5=0.012 mole
let x=Concentration of KMnO4
then 0.24 l * x mole/l =0.012 mole
x=0.05 mole/l.
4.How many grams of calcium chloride can be prepared if a student uses 500g if calcium carbonate and 2.00L of a 4.00 Mol/L sodium chloride solution (Hydrochloric acid) in the reaction.??
CaCO3 + 2HCl -> CaCl2 + H2O
Molecular weight of calcium carbonate
= 40+12+48 = 100
500 g. = 500/100= 5 moles
HCl required for complete reaction with calcium carbonate
= 5*2/1=10 moles
Actual quantity of HCl used
=2 l. * 4 mol/l
=8 moles
Thus number of moles of CaCl2 produced
= 5*8/10=4
Weight of CaCl2 produced
= 4*(40+2*35.5)=444 g.
Thanks man!!!could u also help me out on #8 thank you sooo much!!
I would like to, but I am not sure under what conditions the reaction indicated in #8 will take place. It involves the dissociation of ammonia into nitrogen and hydrogen. Do you have the source of the equation?
tht is the only info that is given .i don't know wat to do//u helped me on a question yestesday #3 i think u forgot to use STP for that one u, can go and check it over///thanks
#3.
The problem is the question.
Water vapour calculated was for STP.
However, if it was pure water vapour, it would condense to form water, and does not exist any more in vapour form.
On the other hand, if this amount is dissipated in the air as humidity, it can exist at STP.
8. 2(NH4)3PO4-->3N2+O2+12H2+2PO3
I believe the equation is not exact or not complete.
The left hand side is ammonium triphosphate.
The right hand side is not clear to me if it is nitrogen, oxygen, H+ and PO33+ ions, but they do not balance.
they asked to name the products and type of reaction
tht is all that is given..i don't know
they asked to name the products and type of reaction//thanks
5. To find the concentration of the KMnO4 solution required to oxidize 2.7g of oxalic acid, we can use the stoichiometry of the balanced chemical equation.
First, let's calculate the molar mass of oxalic acid:
C2H2O4 = (2 * atomic mass of carbon) + (2 * atomic mass of hydrogen) + (4 * atomic mass of oxygen)
= (2 * 12.01 g/mol) + (2 * 1.01 g/mol) + (4 * 16.00 g/mol)
= 90.03 g/mol
Next, we need to convert the mass of oxalic acid to moles:
moles of oxalic acid = mass of oxalic acid / molar mass of oxalic acid
= 2.7 g / 90.03 g/mol
≈ 0.03 mol
Now, let's look at the balanced chemical equation:
5H2C2O4 (aq) + 6H+ + 2MnO4- (aq) -> 10CO2 (g) + 2 Mn2+(aq) + 8H2O (l)
According to the stoichiometry, it takes 2 moles of KMnO4 to react with 5 moles of oxalic acid. Therefore, the required moles of KMnO4 can be calculated as:
moles of KMnO4 = (moles of oxalic acid) * (2 moles of KMnO4 / 5 moles of oxalic acid)
= 0.03 mol * (2/5)
= 0.012 mol
Finally, we need to calculate the concentration of the KMnO4 solution:
Concentration = (moles of KMnO4) / (volume of solution in liters)
= 0.012 mol / 0.24 L
= 0.05 M
Therefore, the concentration of the KMnO4 solution required to oxidize 2.7g of oxalic acid is 0.05 M.
4. To find the amount of calcium chloride that can be prepared using 500g of calcium carbonate and 2.00L of a 4.00 Mol/L sodium chloride solution, we need to use the stoichiometry of the balanced chemical equation.
First, calculate the molar mass of calcium carbonate (CaCO3):
Molar mass of CaCO3 = (atomic mass of calcium) + (atomic mass of carbon) + (3 * atomic mass of oxygen)
= 40.08 g/mol + 12.01 g/mol + (3 * 16.00 g/mol)
= 100.09 g/mol
Next, convert the mass of calcium carbonate to moles:
moles of CaCO3 = mass of CaCO3 / molar mass of CaCO3
= 500 g / 100.09 g/mol
= 4.998 mol
Now, let's look at the balanced chemical equation:
CaCO3 + 2HCl -> CaCl2 + H2O
According to the stoichiometry, 1 mole of calcium carbonate reacts with 2 moles of hydrochloric acid to produce 1 mole of calcium chloride. Therefore, the moles of calcium chloride that can be prepared can be calculated as:
moles of CaCl2 = (moles of CaCO3) * (1 mole of CaCl2 / 1 mole of CaCO3)
= 4.998 mol * 1
= 4.998 mol
Finally, to find the mass of calcium chloride, we need to use its molar mass:
molar mass of CaCl2 = (atomic mass of calcium) + (2 * atomic mass of chlorine)
= 40.08 g/mol + (2 * 35.45 g/mol)
= 110.98 g/mol
mass of CaCl2 = moles of CaCl2 * molar mass of CaCl2
= 4.998 mol * 110.98 g/mol
= 554.62 g
Therefore, the student can prepare 554.62 grams of calcium chloride.
8. Let's address each part of the question individually:
a) The products of the reaction 2(NH4)3PO4 are:
- 3 moles of nitrogen gas (N2)
- 1 mole of oxygen gas (O2)
- 12 moles of hydrogen gas (H2)
- 2 moles of phosphorus oxide (PO3)
b) The reactant in the equation is 2(NH4)3PO4, which represents a compound containing ammonium ions (NH4+) and phosphate ions (PO4^-3).
c) The type of reaction is a decomposition reaction. In a decomposition reaction, a compound breaks down into two or more simpler substances.