Respuesta :
Answer:
a)135.9 grams
b) 10 grams
c) 8.83g
d) 88.3g
Explanation:
the balanced reaction shows that four molecules of iron (II) dichromate will react with eight molecules of potassium carbonate and one molecule of oxygen to give two molecules of ferric oxide, eight molecules of potassium chormate and carbon dioxide each.
a) How many grams of iron (II) dichromate are required to produce 44.0 grams of carbon dioxide?
The molar mass of carbon dioxide = 44 g/mol
The molar mass of iron (II) dichromate = 271.8 g/mol
Thus as we need one mole of carbon dioxide we need to have half moles of iron (II) dichromate = 0.5 X 271.8 = 135.9 grams
b) How many grams of oxygen gas are required to produce 100.0 grams of ferric oxide?
The molar mass of ferric oxide =160 g/mol
the molar mass of oxygen = 32 g/mol
two moles of ferric oxide are obtained from one mole of oxygen
thus
320g of ferric oxide from 32 grams of oxygen
for 100g of ferric oxide the mass of oxygen required = 10 grams
c) If 300.0 grams of iron (II) dichromate react, how many grams of oxygen gas will be consumed?
The molar mass of iron (II) dichromate = 271.8 g/mol
the molar mass of oxygen = 32 g/mol
With four moles of iron (II) dichromate one mole of oxygen reacts
thus 4 X 271.8 grams of iron (II) dichromate reacts with = 32g of oxygen
hence 300 grams will react with =
[tex]\frac{32X300}{4X271.8}=8.83g[/tex]
d) How many grams of iron (III) oxide will be produced from 300.0 grams of ferrous dichromate?
The molar mass of iron (II) dichromate = 271.8 g/mol
The molar mass of ferric oxide =160 g/mol
from 4 moles of iron dichromate two moles of ferric oxides are formed
thus
4 X 271.8 grams of iron dichromate will give = 2X160 g of ferric oxide
300 grams will give =
[tex]\frac{2X160X300}{4X271.8}=88.3g[/tex]
