How many grams of potassium oxide would be needed to produce 5.00 g of potassium nitrate (KNO3) ?

Mass = 2.355 g

Explanation:

Given data:

Mass of K₂O needed = ?

Mass of KNO₃ produced = 5.00 g

Solution:

Solution:

K₂O + 2HNO₃     →     2KNO₃ + H₂O

Number of moles of KNO₃:

Number of moles = mass/molar mass

Number of moles = 5.00 g/ 101.1 g/mol

Number of moles = 0.05 mol

now we will compare the moles of KNO₃ and K₂O.

                 KNO₃           :             K₂O

                    2                :               1

                   0.05           :           1/2×0.05 = 0.025 mol

Mass of potassium oxide needed in gram:

Mass =  number of moles × molar mass

Mass = 0.025 mol × 94.2 g/mol

Mass = 2.355 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

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2