If 62.8 g ca react with 92.3 g hcl according to the reaction below, how many grams of hydrogen gas will be produced, and how many grams of the excess reactant will be left over? unbalanced equation: ca + hcl → cacl2 + h2

2.55 g of hydrogen

12.17 g calcium.

to nearest hundredth.

Explanation:

The balanced equation is:

Ca + 2HCl --->   CaCl2 + H2

Using the atomic masses

40.078 g Ca react with 72.916 g of HCl to give 2.016 g HCl

The ratio of Ca to HCl in the above is 1 to 1.81935

so 62.8 g Ca reacts with 62.8 * 1.81935 = 114.245 g HCl

so there is excess of Ca in the  given weights.

Therefore the mass of Hydrogen produced

=  (2.016 / 72.916) * 92.3

= 2.552 g of hydrogen gas.

The mass of calcium required to produce 2.552 g of hydrogen is:

(2.552 /  2.016) * 40.078

= 50.73 g

So the excess of calcium is 62.8 - 50.73

= 12.17 g.

b. th3 n4zc4 pl4t3 15 c0nv3rg1ng w1th the s0uth 4m3ric4n pl4t3.

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