Balance the following fe(s) +H_2O(g) rightarrow fe_3o_4 (s) + H_2(g) 2, 4, 1, 4

Question

Balance the following fe(s) +H_2O(g) rightarrow fe_3o_4 (s) + H_2(g) 2, 4, 1, 4 3, 2, 1, 4 4, 2, 1, 5 3, 4, 1, 4 In the reaction, 2 H202 rightarrow 2 H_20 + 0_2, what is the number of moles of 0_2 produced by the decomposition of 32g of hydrogen peroxide, H_202- 34 0.470 0.941 1.88 In the above question #7, what is the estimated percent yield if you isolated only 8g of 02? 53% 58% 60% 64% In the reaction, 2 A1 + 3 Cl2 rightarrow Al_2Cl_6 If you mix 5.40 g of A1 with 8.10g of Cl_2 What mass of Al_2Cl_6 can form? 8.1g 9.6g 10.lg 11.lg

Explanation / Answer

5. Balanced equation coefficients,

a. 2,4,1,4

6. In the reaction,

moles of Zn = 7 g/65.38 g/mol = 0.11 mol

moles of HCl = 0.1 mol

If all of Zn is consumed, we would need = 0.11 x 2 = 0.22 mol HCl

Since mols of HCl available is less, the limiting reactant would be,

c. HCl

7. moles of H2O2 = 32 g/34.0147 g/mol = 0.94 mols

2 moles of H2O2 gives 1 mole of O2

So moles of O2 produced would be,

b. 0.470

8. Percent yield = 8 x 100/(0.47 mol x 32 g/mol) = 53%

Answer : a. 53%

9. moles of Al = 5.40 g/26.98 g/mol = 0.20 mols

moles of Cl2 = 8.10 g/70.906 g/mol = 0.114 mols

Is all of Al is consumed, we would need = 0.20 x 3/2 = 0.3 mols of Cl2

Since moles of Cl2 is les than required, this is limiting reagent

Mass of Al2Cl6 formed =0.114 mol x 266.68 g/mol/3 = 10.1 g

Answer : c. 10.1 g

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