_____ NaOH + _____ Al rightarrow _____ Al_2O_3 + _____ H_2 + _____ Na_2O _____ K

Question

_____ NaOH + _____ Al rightarrow _____ Al_2O_3 + _____ H_2 + _____ Na_2O _____ KMnO_4 rightarrow _____ K_2O + _____ MnO + _____ O_2 _____ H_2O + _____ O_2 + _____ As rightarrow _____ HAsO_2 _____ Au_2O_3 rightarrow _____ Au + _____ O_2 Calculate: The mass (in grams) of 1.058 mole of H_2O. The number of atoms in 0.750 mole of Cu. The mass of 1.2046 times 10^23 atoms of Fe Calculate the molar masses (in g/mol) for: Cesium Chloride Na_2SO_4 Given the reaction represented by the balanced equation: CH_4 (g) + 3Cl_2 (g) rightarrow 3 HCl (g) + CHCl_3 (g) Calculate the number of grams HCl that could be produced by mixing 105 g Cl_2 excess CH_4. If 10.0 g HCl were actually produced, calculate the % yield.

Explanation / Answer

Quesiton 1-4 have no instructions

Question 5.

mass of H2O in n = 1.058 mol

1 mol of H2O = 18 g /mol

mass =mol*MW = 1.058 *18 = 19.044 g of H2O

b)

number of atoms in Cu --> 0.75 mol

1 mol = 6.022*10^23 atoms

0.75 mol --> (0.75)* (6.022*10^23 ) = 4.5165*10^23 atoms of copper

c)

1 mol of Fe = 55.8 g /mol

find moles in

1.2046*10^23 atoms

1 mol = 6.022*10^23

moles = (1.2046*10^23)/(6.022*10^23) = 0.20 moles

now

mass = mol*MW = 0.20*55.8 = 11.16 g of Fe

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