Answer:
First, balance the half-reactions
Second, equalize the electrons
Third,add two reaction equations to get final answer
Explanation:
For example
H₂C₂0₄ + MnO⁻₄ ---------->CO₂+Mn²⁺
(i) Balancing the half reactions
H₂C₂O₄-------->2CO₂+2H⁺+2e⁻
5e⁻ +8H⁺+MnO₄⁻----------->Mn²⁺+4H₂O
(ii)
Equalizing the electrons
5H₂C₂O₄--------->10CO₂+10H⁺+10e⁻ ---here there is a factor of 5
10e⁻+16H⁺+2MnO₄⁻--------->2Mn²⁺+8H₂O -----here there is a factor of 2
(iii)
Add the two where electrons and some Hydrogen ions will cancel out
5H₂C₂O₄+6H⁺+2MnO₄⁻---->10CO₂+2Mn²⁺+8H₂O
Camels lose less water through their urine and feces than many other mammals. Their kidneys concentrate water heavily, leading to salty urine. The intestines also reabsorb water from intestinal material as it is digested, so dry feces are produced.
Answer:
Theoretical yield of the reaction = 34 g
Excess reactant is hydrogen
Limiting reactant is nitrogen
Explanation:
Given there is 100 g of nitrogen and 100 g of hydrogen
Number of moles of nitrogen = 100 ÷ 28 = 3·57
Number of moles of hydrogen = 100 ÷ 2 = 50
Reaction between nitrogen and hydrogen yields ammonia according to the following chemical equation
N2 + 3H2 → 2NH3
From the above chemical equation for every mole of nitrogen that reacts, 3 moles of hydrogen will be required and 2 moles of ammonia will be formed
Now we have 3·57 moles of nitrogen and therefore we require 3 × 3·57 moles of hydrogen
⇒ We require 10·71 moles of hydrogen
But we have 50 moles of hydrogen
∴ Limiting reactant is nitrogen and excess reactant is hydrogen
From the balanced chemical equation the yield will be 2 × 3·57 moles of ammonia
Molecular weight of ammonia = 17 g
∴ Theoretical yield of the reaction = 2 × 3·57 × 17 = 121·38 g
1) Balanced chemical equation:
2SO2 (g) + O2 (g) -> 2SO3 (l)
2) Molar ratios
2 mol SO2 : 1 mol O2 : 2 mol SO3
3) Convert 6.00 g O2 to moles
number of moles = mass in grams / molar mass
number of moles = 6.00 g / 32 g/mol = 0.1875 mol O2.
4) Use proportions with the molar ratios
=> 2 moles SO2 / 1 mol O2 = x / 0.1875 mol O2
=> x = 0.1875 mol O2 * 2 mol SO2 / 1 mol O2 = 0.375 mol SO2.
5) Convert 0.375 mol SO2 to grams
mass in grams = number of moles * molar mass
molar mass SO2 = 32 g/mol + 2*16 g/mol = 64 g/mol
=> mass SO2 = 0.375 mol * 64 g / mol = 24.0 g
Answer: 24.0 g of SO2 are needed to react completely with 6.00 g O2.