Determine the type of chemical reaction, find the limiting reagent, determine the amount of product in grams, and the amount of grams of excess reagent.
Chromium metal is heated with oxygen to give chromium (III) oxide. Start with 38.5 moles chromium metal and unlimited oxygen gas.
The equation is 4Cr(s) + 3O2(g) --> 2Cr2(O2)3(s) and it is a synthesis reaction. I am not sure how to figure out the rest.
I tried something for this. could you please check?
Limiting reagent is Cr because O2 is unlimited.
For every 4 moles of Cr, there's 3 moles of O2.
38.5molesCr (3molesO2 / 4molesCr) = 28.9 moles O2.
Molar Mass of O2 is 32 so 1moleO2 = 32gO2.
28.9molesO2 (32gO2 / 1moleO2) = 924gO2.
38.5molesCr (52gCr / 1moleCr) = 2002gCr
Mass of Cr2(O2)3 is 924 + 2002 = 2926.
I don't know if I answered the last question and I don't know if this is right. Please let me know. Thanks!
To determine the type of chemical reaction, you correctly identified it as a synthesis reaction. A synthesis reaction is a type of chemical reaction where two or more substances combine to form a more complex product.
To find the limiting reagent, we need to compare the stoichiometry of the reactants in the balanced equation with the given amounts.
In the balanced equation:
4 moles of chromium (Cr) react with 3 moles of oxygen (O2) to produce 2 moles of chromium(III) oxide (Cr2O3).
Given:
38.5 moles of chromium metal (Cr)
Unlimited oxygen gas (O2)
To determine the limiting reagent, we need to calculate the moles of product that can be formed using both reactants.
Moles of chromium (Cr): 38.5 moles
According to the equation, 4 moles of chromium (Cr) produce 2 moles of chromium(III) oxide (Cr2O3).
So, the moles of chromium(III) oxide that can be produced from the given amount of chromium is:
38.5 moles Cr x (2 moles Cr2O3 / 4 moles Cr) = 19.25 moles Cr2O3
Since you have an unlimited amount of oxygen gas, it is not the limiting reagent in this case. The limiting reagent is the reactant that gets completely consumed and limits the amount of product formed.
The amount of product in grams can be determined by multiplying the moles of the limiting reagent by its molar mass. In this case, the limiting reagent is chromium (Cr).
Molar mass of chromium(III) oxide (Cr2O3): 2(52 g/mol Cr) + 3(16 g/mol O) = 152 g/mol Cr2O3
Grams of chromium (III) oxide (Cr2O3) produced:
19.25 moles Cr2O3 x (152 g Cr2O3 / 1 mol Cr2O3) = 2,930 g Cr2O3
To find the amount of grams of excess reagent, subtract the grams of limiting reagent (chromium(III) oxide) produced from the total grams of chromium (Cr) used.
Grams of excess reagent (chromium) = (38.5 moles Cr x 52 g/mol Cr) - 2,930 g Cr2O3
Simplify:
Grams of excess reagent (chromium) = 2,002 g Chromium
Please note that this calculation assumes that the reaction goes to completion and that there are no other limiting factors.