Question

< CH7HW 11/07/19 12 out of 23 15 Question (1 point) These questions are based on the following balanced equation: 2H, +0, 2H,0 If you start with 6.833 g of H2 and 3.095 g of Oz, find the following: 1st attempt Part 1 (0.3 point) With excess O2, what mass of H2O would be produced by the H2? g H20 < 15/20 > acer esc

Answer 1

1)

Molar mass of H2 = 2.016 g/mol

mass of H2 = 6.833 g
mol of H2 = (mass)/(molar mass)
= 6.833/2.016
= 3.389 mol

According to balanced equation
mol of H2O formed = moles of H2
= 3.389 mol


Molar mass of H2O,
MM = 2*MM(H) + 1*MM(O)
= 2*1.008 + 1*16.0
= 18.016 g/mol


mass of H2O = number of mol * molar mass
= 3.389*18.02
= 61.06 g
Answer: 61.06 g

2)

Molar mass of O2 = 32 g/mol

mass of O2 = 3.095 g
mol of O2 = (mass)/(molar mass)
= 3.095/32
= 9.672*10^-2 mol

According to balanced equation
mol of H2O formed = (2/1)* moles of O2
= (2/1)*9.672*10^-2
= 0.1934 mol


Molar mass of H2O,
MM = 2*MM(H) + 1*MM(O)
= 2*1.008 + 1*16.0
= 18.016 g/mol


mass of H2O = number of mol * molar mass
= 0.1934*18.02
= 3.485 g
Answer: 3.485 g

3)
Less amount of H2O is formed in case of O2
So, O2 is limiting reagent
Answer: O2