When 29.7 g of HCI and 28.4 g of CaCO_2 are mixed both reactants are completely

Question

When 29.7 g of HCI and 28.4 g of CaCO_2 are mixed both reactants are completely consumed. What mass of CO_2 gas is produced if the mass of he other producers formed is 36.6 g? (A) 85.7 g (B) 15.9 (C) 12.5 g (D) 8.20 g What amount of the excess reagent remains when 0.500 mol Li reacts with 0.550 mol N_2? 6Li(s) + N_2(g) rightarrow 2LiN(s) (A) 0.0833 mol N_2 (B) 0.150 mol Li (C) 0.267 mol N_2 (D) 0.417 mol Li when 8.70 g AI reacted with excess O_2, 12.8 g of AI_3O_3 was obtained. What was the percent yield? 4AI + 3O_2 rightarrow 2AI_2O_3 (A) 16.4% (B) 38.9% (C) 68.0% (D) 78.0% What is the maximum number of water molecules produced if 0.34 mol of propane, C_3H_2, combustion in excess oxygen? C_3H_2(g) + 5O_2(g) rightarrow 3CO_2(g) + 4H_2O(g) (A) 2.4 times 10^34 molecules (B) 8.2 times 10^23 molecules (C) 2.1 times 10^23 molecules (D) 5.1 times 10^22 molecules What is the maximum mass of NO_2 that is formed when 7.58 g of N_2O_3 are consumed? 2N_2O_3(g) rightarrow 4NO_2(g) + O_2(g) (A) 1.6 g (B) 3.2 g (C) 6.4 g (D) 15 g What is the oxidation number of chlorine in NaCIO_2? (A) +5 (B) +3 (C) +1 (D) -1 Which reaction will result in a precipitate? (A) KCI(aq) + CaBr_2(aq) rightarrow (B) NH_4NO_2(aq) + HI(aq) rightarrow (C) FeCI_2(aq) + Na_2S(aq) rightarrow (D) H_2SO_4(aq) + NaOH(aq) rightarrow Which is an example of an oxidation-solutions reaction? (A) 2H_2O_2(aq) rightarrow O_2(g) + 2H_2O(l) (B) CaCO_2(s) rightarrow CaO(s) + CO_2(g) (C) NaHCO_2(s) + HCI(aq) rightarrow NaCI(aq) + CO_2(g) + H_2O(l) (D) 2Pb^2+(aq) + C_2O_1^2(aq) + H_2O(l) rightarrow 2PbCrO_4(s) + 2H^+(aq) Assuming volumes are additive, what volume of water must be added to 28.0 mL of 1.25 M FeCI_2 to prepare a 0.650 M FeCI_2 solution? (A) 81.8 mL (B) 53.8 mL (C) 25.8 mL (D) 14.6 mL What is the molar concentration of the chloride ion is a 3.0 M CaCI_2 solution? (A) 1.0 M (B) 1.5 M (C) 3.0 M (D) 6.0 M What mass of LiF si needed to make 500.0 mL of a 0.500 M solution? (A) 25.9 g (B) 13.0 g (C) 6.49 g (D) 0.250 g

Explanation / Answer

Q12

fnid mass of Oxygen in

25.5 g of Al2(CO3)2

so

mol of Al2(CO3)2 --> mass/MW = 25.5/233,9898 = 0.1089791 mol of Al2(CO3)3

then

1 mol of Al2(CO3)2 = 9 mol of O

0.1089791 mol --> 0.1089791 *9 = 0.9808119 mol of O

therefore

mass of O = mol*MW = 0.9808119*16 = 15.69299 g of A

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