4 questions! I put stars next to the answers I think are right.
. In the following reaction, how many moles of FeO are needed to react completely with 1.6 mol of Al?
2Al (s)+ 3FEO (s)-> 3FE (s)+Al2O3 (s)
1.2, 1.8*, 2.4, 3.0
What is the maximum number of grams of PH3 that can form when 12.4 g of phosphorus reacts with 8.0 g of hydrogen to form PH3?
.86 g 13.6 g 90 g 270 g*
Lead oxide can be decomposed by heating. What is the percent yield of the decomposition reaction if 22.3 grams of PbO are heated to give 18.0 g of Pb?
2PBO -> 2PB+ O2
How many grams of CO are needed with an excess of Fe2O3 to produce 27.9 g Fe? Please show work.
no, what is (3/2)*1.6?
2P+H2>>2PH3, no
no work shown on last two.
Ah, so 1 would be 2.4?
As for 2, I've tried it again and I got 90g
To find the number of moles of FeO needed to react completely with 1.6 mol of Al, we can use the balanced equation:
2Al (s) + 3FeO (s) -> 3Fe (s) + Al2O3 (s)
From the balanced equation, we can see that the mole ratio of Al to FeO is 2:3. Therefore, for every 2 moles of Al, we need 3 moles of FeO.
Using this ratio, we can set up a proportion:
(3 moles FeO / 2 moles Al) = (x moles FeO / 1.6 moles Al)
Cross-multiplying and solving for x, we get:
(3 moles FeO * 1.6 moles Al) / 2 moles Al = x moles FeO
Simplifying this equation gives us:
x = 2.4 moles FeO
Therefore, the correct answer is 2.4 moles FeO.
For the second question, we need to determine the maximum number of grams of PH3 that can form when 12.4 g of phosphorus reacts with 8.0 g of hydrogen. The balanced chemical equation for this reaction is:
P4 + 6H2 -> 4PH3
From the balanced equation, we can see that the mole ratio of P4 to PH3 is 1:4. Therefore, for every mole of P4, we get 4 moles of PH3.
First, let's calculate the number of moles of phosphorus present in 12.4 g:
Molar mass of phosphorus (P) = 31.0 g/mol
Number of moles of P = mass of P / molar mass of P
Number of moles of P = 12.4 g / 31.0 g/mol = 0.4 mol P
Now, let's calculate the number of moles of PH3 that can be formed:
Mole ratio of P4 to PH3 = 1:4
Number of moles of PH3 = 4 * number of moles of P
Number of moles of PH3 = 4 * 0.4 mol = 1.6 mol PH3
Finally, let's convert the number of moles of PH3 to grams:
Molar mass of PH3 = 33.97 g/mol
Mass of PH3 = number of moles of PH3 * molar mass of PH3
Mass of PH3 = 1.6 mol * 33.97 g/mol = 54.35 g
Therefore, the maximum number of grams of PH3 that can form is 54.35 g.
For the third question, we need to calculate the percent yield of the decomposition reaction of PbO. The balanced equation for this reaction is:
2PbO -> 2Pb + O2
First, let's determine the number of moles of PbO initially present:
Molar mass of PbO = 223.2 g/mol (using the periodic table)
Number of moles of PbO = mass of PbO / molar mass of PbO
Number of moles of PbO = 22.3 g / 223.2 g/mol = 0.1 mol PbO
The theoretical yield of Pb is the mass that would be obtained if the reaction goes to completion. From the balanced equation, we can see that the mole ratio of PbO to Pb is 2:2 (or 1:1). Therefore, the theoretical yield of Pb is also 0.1 mol.
The actual yield of Pb is given as 18.0 g.
Now, let's calculate the percent yield:
Percent yield = (actual yield / theoretical yield) * 100
Percent yield = (18.0 g / 0.1 mol) * 100
Percent yield = 180%
Therefore, the percent yield of the decomposition reaction is 180%.
Finally, for the fourth question, we need to calculate the mass of CO needed to produce 27.9 g of Fe. The balanced equation for this reaction is:
3CO + Fe2O3 -> 2Fe + 3CO2
From the balanced equation, we can see that the mole ratio of CO to Fe is 3:2. Therefore, for every 3 moles of CO, we get 2 moles of Fe.
First, let's calculate the number of moles of Fe present:
Molar mass of Fe = 55.8 g/mol
Number of moles of Fe = mass of Fe / molar mass of Fe
Number of moles of Fe = 27.9 g / 55.8 g/mol = 0.5 mol Fe
Now, let's calculate the number of moles of CO needed:
Mole ratio of CO to Fe = 3:2
Number of moles of CO = (3/2) * number of moles of Fe
Number of moles of CO = (3/2) * 0.5 mol = 0.75 mol CO
Finally, let's convert the number of moles of CO to grams:
Molar mass of CO = 28.01 g/mol
Mass of CO = number of moles of CO * molar mass of CO
Mass of CO = 0.75 mol * 28.01 g/mol = 21.01 g
Therefore, approximately 21.01 grams of CO are needed.