Mg(s) + 2HCl (aq) > MgCl2 (aq) + H2 You have 6.00 g HCl and 4.8 g Mg.
What is the limiting reaction

HCl is the limiting reactant

Explanation:

To find the limiting reactant of a reaction we must find the moles of each one as follows:

Moles Mg (Molar mass: 24.305g/mol)

4.8g Mg * (qmol / 24.305g) = 0.20 moles Mg

Moles HCl (Molar mass: 36.46g/mol)

6.00g HCl* (1mol / 36.46g) = 0.165 moles HCl

Based on the chemical reaction, for a complete reaction of 0.20 moles of Mg are necessaries:

0.20 moles Mg * (2mol HCl / 1mol Mg) = 0.40 moles HCl

As there are just 0.165 moles of HCl:

i had this on my unit activity, i can you if you are a plato student. are you a plato student?

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