Basic Stoichiometry Part 1: Mole ←→ Mass Conversions
Choose 2: Convert the following number of moles of chemical into its mass in grams.
1. 0.436 moles of (NH4)Cl = g
2. 0.50 moles of Ca(NO3)2 = g
Choose 2: Convert the following masses into their corresponding number of moles.
3. 23.5 g of NaCl = moles
4. 79.9 g of KMnO4 =moles
5) The combustion of a sample of butane, C4H10 (lighter fluid), produced 2.46 grams of water.

1. 23.3 g

2. 82 g

3. 0.402 mol

4. 0.506 mol

Explanation:

We can convert moles (n) to mass (m) using the following expressions that include the molar mass (M).

n = m/M

m = n × M

1. The molar mass of NH₄Cl is 53.49 g/mol.

m = 0.436 mol × 53.49 g/mol = 23.3 g

2. The molar mass of Ca(NO₃)₂ is 164.09 g/mol.

m = 0.50 mol × 164.09 g/mol = 82 g

3. The molar mass of NaCl is 58.44 g/mol.

n = 23.5 g/(58.44 g/mol) = 0.402 mol

4. The molar mass of KMnO₄ is 158.03 g/mol.

n = 79.9 g/(158.03 g/mol) = 0.506 mol

here's the link to your question!

explanation:

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