Answer:
The answer is c
Explanation:
Because all of the Other ones did not make sense
Balance the equation first:
2 Fe+6 HNO3→2 Fe(NO3)3+3H2
Then calculate mass of Iron :
4.5×3.0×3.5 cm3(1 mL1 cm3)(7.87 g Fe1 ml)=371.86 g Fe
Now use Stoichiometry:
371.86 g Fe×(1 mol Fe55.85 g Fe)×(6 mol HNO32 mol Fe)=19.97 mol HNO3
Convert moles of nitric acid to grams
19.97 mol HNO3×(63.01 g HNO31 mol HNO3)=1258.3 g HNO3
Answer:
Mass = 88.12 g
Explanation:
Given data:
Mass of iron oxide = 126 g
Mass of iron formed = ?
Solution:
Chemical equation:
Fe₂O₃ + 3CO → 2Fe + 3CO₂
Number of moles of iron oxide:
Number of moles = mass/molar mass
Number of moles = 126 g/ 159.69 g/mol
Number of moles = 0.789 mol
Now we will compare the moles of iron with iron oxide.
Fe₂O₃ : Fe
1 : 2
0.789 : 2/1×0.789 = 1.578 mol
Mass of iron:
Mass = number of moles ×molar mass
Mass = 1.578 mol × 55.84 g/mol
Mass = 88.12 g
Answer:
26.9 g
81%
Explanation:
The equation of the reaction is;
4 KO2(s) + 2 CO2(g) → 3 O2(g) + 2 K2CO3(s)
Number of moles of KO2= 27.9g/71.1 g/mol = 0.39 moles
4 moles of KO2 yields 2 moles of K2CO3
0.39 moles of KO2 yields 0.39 × 2/4 = 0.195 moles of K2CO3
Number of moles of CO2 = 57g/ 44.01 g/mol = 1.295 moles
2 moles of CO2 yields 2 moles of K2CO3
1.295 moles of CO2 yields 1.295 × 2/2 = 1.295 moles of K2CO3
Hence the limiting reactant is KO2
Theoretical yield = 0.195 moles of K2CO3 × 138.205 g/mol = 26.9 g
Percent yield = actual yield/theoretical yield × 100
Percent yield = 21.8/26.9 × 100
Percent yield = 81%
