Question

1. in the reaction of solid copper with silver nitrate, solid silver and copper(ii) nitrate form. how many grams of solid silver will form if 24.58 g of solid copper are used?

b. what is the mole ratio between the hydrochloric acid and the calcium carbonate?
c. if 0.68 g hydrochloric acid reacts with excess calcium carbonate, what amount (in moles) of calcium chloride will be produced?
of carbon monoxide gas and hydrogen gas to produce octane (c₈h₁₈) and water

Explanation:

Step 1: Write the balanced chemical equation

The reaction between copper (Cu) and silver nitrate ($\ce{AgNO_3}$) is a single - displacement reaction. The balanced equation is:
$\ce{Cu(s) + 2AgNO_3(aq) -> Cu(NO_3)_2(aq) + 2Ag(s)}$

Step 2: Calculate the moles of copper

The molar mass of copper (Cu) is $M_{Cu}=63.55\space g/mol$. Given the mass of copper $m_{Cu} = 24.58\space g$.
Using the formula $n=\frac{m}{M}$, where $n$ is the number of moles, $m$ is the mass, and $M$ is the molar mass.
$n_{Cu}=\frac{24.58\space g}{63.55\space g/mol}\approx0.3868\space mol$

Step 3: Determine the moles of silver from the mole ratio

From the balanced equation, the mole ratio of $\ce{Cu}$ to $\ce{Ag}$ is $1:2$. So, $n_{Ag}=2\times n_{Cu}$
$n_{Ag}=2\times0.3868\space mol = 0.7736\space mol$

Step 4: Calculate the mass of silver

The molar mass of silver (Ag) is $M_{Ag}=107.87\space g/mol$.
Using the formula $m = n\times M$, we get:
$m_{Ag}=n_{Ag}\times M_{Ag}=0.7736\space mol\times107.87\space g/mol\approx83.45\space g$

Answer:

The mass of solid silver formed is approximately $\boldsymbol{83.45\space g}$