Abalanced chemical equation used to prepare ammonium carbonate, (nh4)2co3 , is: 2 n h 3 ( g ) + c o 2 ( g ) + h 2 o ( l ) ⟶ ( n h 4 ) 2 c o 3 which choice of reactant quantities shown below would result in the greatest amount of ammonium carbonate being formed?

This a simple stoichiometry problem using the ideal gas law.  

First take the grams of ammonium carbonate and convert it to moles using its molar mass and dividing. 11.9 g/96.0932 g/mol= .12384 mol  

Now use a molar conversion using the balanced equation,  

1 mol (NH4)2CO3 ---> 4 mol gas formed (2 mol NH3 + 1 mol CO2 + 1 mol H2O) = .12384 x 4 = .49535 mol gas  

PV=nRT  

V=nRT/P= .49535mol (.08206 Lxatm/molxK) (296K)/ (1.03 atm)=11.682 L

answer; blank (1); /// (2)///(3)///highblank(4)//low;

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