Question

49. Calculate the mass of water produced when 625 of methane, CH4, reacts with an excess of oxygen in the following unbalanced reaction. CH4%)+18) + CO2g)+H2Og) a. 7.02 g H20 b. 2.25 x 10² g H20 c. 14.0 g H20 d. 0.347 g H20 c. 0.78 g H20 50. In the reaction below, how many molecules of hydrogen are required to react with 9.79 g of nitrogen? 3H,@) + N2) 2NH3) 1.2.10 x 1023 molecules hydrogen b. 3.16 x 1023 molecules hydrogen c. 1.40 x 1023 molecules hydrogen d. 6.31 x 1023 molecules hydrogen c. None of these 51. How many atoms of aluminum can be produced by the decomposition of 13.3 g of aluminum oxide? (Hint: Write down the balanced equation before solving.) a. 3.14 x 1023 aluminum atoms b. 1.57 x 1023 aluminum atoms c. 7.86 x 1022 aluminum atoms d. 3.93 x 10" aluminum atoms e. None of these 52. For the reaction of C2H4(g) with O2(g) to form CO2(g) and H2O(g), what number of moles of CO2 can be produced by the reaction of 0.480 mol C2H4 and 1.17 mol O2? a. 1.76 mol b. 0.960 mol c. 0.780 mol d. 1.17 mol e. none of these

Answer 1

49)

Molar mass of CH4,

MM = 1*MM(C) + 4*MM(H)

= 1*12.01 + 4*1.008

= 16.042 g/mol

mass(CH4)= 6.25 g

use:

number of mol of CH4,

n = mass of CH4/molar mass of CH4

=(6.25 g)/(16.04 g/mol)

= 0.3896 mol

Balanced chemical equation is:

CH4 + 2 O2 ---> 2 H2O + CO2

Molar mass of H2O,

MM = 2*MM(H) + 1*MM(O)

= 2*1.008 + 1*16.0

= 18.016 g/mol

According to balanced equation

mol of H2O formed = (2/1)* moles of CH4

= (2/1)*0.3896

= 0.7792 mol

use:

mass of H2O = number of mol * molar mass

= 0.7792*18.02

= 14.04 g

Answer: 14.0 g

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