If you burn 57.9 g of hydrogen and produce 517 g of water, how much oxygen reacted?

In order to make water (H2O) out of hydrogen (H2), you also need OXYGEN (O2) gas.  So, first write a correctly balanced equation for this reaction.

2H2(g) + O2(g) ==> 2H2O

Now, convert 59.5 g of H2 to moles, then figure out moles of H2O and then moles of O2 used to do this.  

moles H2 = 59.5 g H2 x 1 mole/2 g = 29.8 moles H2

moles H2O formed = 531 g H2O x 1 mole/18 g = 29.5 moles H2O

So, the mole ratio of H2 to H2O is 1:1 meaning that the moles of O2 used will be 1/2 that amount.  This is the case because from the balanced equation you see that 1 O2 reacts with 2 H2 and produces 2 H2O. 

Amount of O2 used = 29.8/2 = 14.9 moles

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