42.6 g Cu are combined with 84.0 g HNO3 according to reaction:
3Cu + 8HNO3 _> 3Cu(NO3)2 + 2NO + 4H2O
What reagent is limiting and how many grams of Cu(NO3)2 are produced?

Mass of Cu(NO₃)₂ produced = 125.67 g

limiting reactant = Cu

Explanation:

Given data:

Mass of Cu = 42.6 g

Mass of HNO₃ = 84.0 g

What is limiting reactant = ?

Mass of Cu(NO₃)₂ produced = ?

Solution:

Chemical equation:

3Cu + 8HNO₃    →     3Cu(NO₃)₂ + 4H₂O + 2NO

Number of moles of Cu:

Number of moles = mass/ molar mass

Number of moles = 42.6 g/63.546 g/mol

Number of moles = 0.67 mol

Number of moles of HNO₃:

Number of moles = mass/ molar mass

Number of moles = 84 g/63.01 g/mol

Number of moles = 1.3 mol

Now we will compare the moles of Cu(NO₃)₂ with HNO₃ and Cu from balance chemical equation.

                    HNO₃         :        Cu(NO₃)₂

                       8              :         3

                      1.3             :       3/8×1.3 = 6

                    Cu              :           Cu(NO₃)₂

                      3               :              3

                      0.67          :           0.67

Number of moles of Cu(NO₃)₂ produced by Cu are less so Cu will limiting reactant.

Mass of Cu(NO₃)₂:

Mass = number of moles × molar mass

Mass = 0.67 mol × 187.56 g/mol

Mass = 125.67 g

answer: the figure is attached herewith.

step-by-step explanation:   we are given to sketch a unit circle and label the terminal points corresponding to

in the attached figure, we have drawn a unit circle, with centre at o(0,0) and radius 1 unit. we have labelled the points a, b, c, d and e corresponding to the angles below

∠aox = 60°,

∠box = 120°,

∠cox = 180°,

∠dox = 240°,

∠eox = 300°.

and the coordinates are a(0.5,1), b(-0.5,1), c(-1,0), d(-0.5,-1) and e(0.5,-1).

thus, the answer is complete.

i believe it is 40–40–20%

explanation: