According to the following reaction, how many grams of sulfur are formed when 37.4 g of water are formed? 2 H2S(g) + SO2(g) → 3 S(s) + 2H2O(l)

Mass = 100.8 g

Explanation:

Given data:

Mass of sulfur form = ?

Mass of water formed = 37.4 g

Solution:

Chemical equation:

2H₂S + SO₂      →      3S + 2H₂O

Number of moles of water:

Number of moles = mass/molar mass

Number of moles = 37.4 g/ 18 g/mol

Number of moles = 2.1 mol

Now we will compare the moles of water with sulfur.

                   H₂O        :         S

                      2          :         3

                  2.1           :     3/2×2.1 = 3.15 mol

Mass of sulfur:

Mass = number of moles × molar mass

Mass = 3.15 mol ×  32 g/mol

Mass = 100.8 g

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

did anyone actually do this? like if anyone has submitted this i rly need .