check answer please
10. 2 Ca3(PO4)2(s) + 6 SiO2(g) + 10 C(s) ---> P4(s) + 6 CaSiO3(s) + 10 CO(g)
If 39.3 g of Ca3(PO4)2, 24.4 g of SiO2 and 8.00g of C are available, find the limiting reagent.
n Ca3(PO4)2= 0.127 moles
n SiO2= 0.135 moles
n C= 0.133 moles
My final answer is Ca3(PO4)2 is the limiting reagent.
If the conversion of Iron ore to iron is represented by Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g) produces 147 g of iron from 500 g of iron ore, what is the percent yield?
Fe2O3(s) + 3CO(g) --> 2Fe(s) + 3CO2(g)
that tells you 1 mole Fe2O3 = 160g gives 2 moles Fe =112g
Use simple proportion
500g Fe2O3 gives 112/160 x 500= 350g
% yield = actual amount / calculated amount x 100 = 147/350 x 100 = 42%
I don't agree with the limiting reagent I get different values for mols SiO2 and mols C. I agree with mols Ca3(PO4)2.
I agree with the 42$ percent yield.
Convert your given mass values to moles then divide by the respective coefficients of the balanced equation. The smallest value is the limiting reactant.
The calculations for both problems are correct.
For the first problem, you correctly found the number of moles for each reactant. From the given amounts, the number of moles for Ca3(PO4)2 is 0.127 moles, for SiO2 is 0.135 moles, and for C is 0.133 moles. Since the stoichiometric ratio of Ca3(PO4)2 to P4 is 2:1, it means that you only need 0.064 moles of Ca3(PO4)2 for the reaction. Since you have more than that amount, Ca3(PO4)2 is not the limiting reagent.
For the second problem, you correctly used the stoichiometric ratio to convert the given mass of Fe2O3 to the expected mass of Fe produced. You then used that calculated amount (350g) to find the percent yield. The percent yield is determined by dividing the actual amount of Fe produced (147g) by the calculated amount (350g) and multiplying by 100. Your calculation gives a percent yield of 42%.
For the first question:
To determine the limiting reagent, we need to compare the moles of each reactant to the stoichiometric ratios in the balanced equation.
Given:
n(Ca3(PO4)2) = 0.127 moles
n(SiO2) = 0.135 moles
n(C) = 0.133 moles
Now, let's find the stoichiometric ratios:
From the balanced equation:
2 Ca3(PO4)2(s) + 6 SiO2(g) + 10 C(s) ---> P4(s) + 6 CaSiO3(s) + 10 CO(g)
The stoichiometric ratio between Ca3(PO4)2 and P4 is 2:1, meaning 1 mole of Ca3(PO4)2 will produce 1 mole of P4.
To find the maximum number of moles of P4 that can be produced, we multiply the moles of Ca3(PO4)2 by the stoichiometric ratio:
0.127 moles of Ca3(PO4)2 * (1 mole of P4 / 2 moles of Ca3(PO4)2) = 0.0635 moles of P4
Similarly, for SiO2 and C:
The stoichiometric ratio between SiO2 and P4 is 6:1, meaning 1 mole of SiO2 will produce 1/6 moles of P4.
0.135 moles of SiO2 * (1/6 moles of P4 / 1 mole of SiO2) = 0.0225 moles of P4
The stoichiometric ratio between C and P4 is 10:1, meaning 1 mole of C will produce 1/10 moles of P4.
0.133 moles of C * (1/10 moles of P4 / 1 mole of C) = 0.0133 moles of P4
Comparing the moles of P4 produced from each reactant, we see that Ca3(PO4)2 produces the highest number of moles of P4 (0.0635 moles), meaning it is the limiting reagent. Therefore, your answer is correct that Ca3(PO4)2 is the limiting reagent.
For the second question:
To find the percent yield, we need to compare the actual amount of product obtained to the theoretically calculated amount of product.
Given:
Mass of iron produced = 147 g
Mass of iron ore used = 500 g
To calculate the theoretical amount of iron that should be produced, we can use the stoichiometric ratios from the balanced equation:
Fe2O3(s) + 3CO(g) --> 2Fe(s) + 3CO2(g)
From the equation, we know that 1 mole of Fe2O3 reacts to produce 2 moles of Fe.
So, the theoretical amount of Fe that can be produced from 500 g of Fe2O3 can be calculated as follows:
1 mole of Fe2O3 reacts with 2 moles of Fe, which is equivalent to 160 g of Fe2O3 reacting to produce 2 * 55.85 g (the molar mass of Fe)
Therefore, 500 g of Fe2O3 will theoretically produce (2 * 55.85 g / 160 g) * 500 g = 348.13 g of Fe.
Now, we can calculate the percent yield using the formula:
% yield = (actual amount / calculated amount) * 100
% yield = (147 g / 348.13 g) * 100 = 42.27%
So, your answer of 42% for the percent yield is correct.