Let's go through each problem step by step.
1. Reaction between copper oxide and carbon
Given reaction: \[ 2 \text{CuO}(s) + \text{C}(s) \rightarrow 2 \text{Cu}(s) + \text{CO}_2(g) \]
Molar masses:
- Cu (Copper) = 64 g/mol
- O (Oxygen) = 16 g/mol
- CuO = 64 + 16 = 80 g/mol
- C (Carbon) = 12 g/mol
a. Calculate the number of moles in 4.0 g of CuO.
\[ \text{Moles of CuO} = \frac{\text{mass}}{\text{molar mass}} = \frac{4.0 \text{ g}}{80 \text{ g/mol}} = 0.050 \text{ moles} \]
b. Calculate the number of moles in 1.2 g of C.
\[ \text{Moles of C} = \frac{1.2 \text{ g}}{12 \text{ g/mol}} = 0.1 \text{ moles} \]
c. Identify the limiting reactant.
The balanced equation shows that:
- 2 moles of CuO react with 1 mole of C.
From the moles calculated:
- Moles of CuO = 0.050 moles
- Moles of C = 0.1 moles
Using the stoichiometry from the balanced equation: For every 1 mole of C, we need 2 moles of CuO. Therefore, for 0.1 moles of C, we need: \[ 0.1 \text{ moles C} \times 2 = 0.2 \text{ moles CuO} \]
Since we only have 0.050 moles of CuO (which is less than 0.2 moles), CuO is the limiting reactant.
Limit Reactant: CuO (Copper oxide)
d. What mass of Cu would you expect to make?
From the balanced equation, when 2 moles of CuO produce 2 moles of Cu, it implies a 1:1 ratio for CuO to Cu.
Since we have 0.050 moles of CuO, it will produce 0.050 moles of Cu.
To find the mass of Cu produced: \[ \text{Mass of Cu} = \text{moles} \times \text{molar mass} = 0.050 \text{ moles} \times 64 \text{ g/mol} = 3.2 \text{ g} \]
2. Reaction between zinc carbonate and hydrochloric acid
Given reaction: \[ \text{ZnCO}_3(s) + 2 \text{HCl}(aq) \rightarrow \text{ZnCl}_2(aq) + \text{H}_2O(l) + \text{CO}_2(g) \]
Molar masses:
- Zn (Zinc) = 65.4 g/mol
- C (Carbon) = 12 g/mol
- O (Oxygen) = 16 g/mol
- H (Hydrogen) = 1 g/mol
- ZnCO3 = 65.4 + 12 + (16 \times 3) = 65.4 + 12 + 48 = 125.4 g/mol
- HCl = 1 + 35.5 = 36.5 g/mol
a. Which reactant is in excess? Show your reasoning.
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Calculate moles of ZnCO$_3$: \[ \text{Moles of ZnCO}_3 = \frac{6.25 \text{ g}}{125.4 \text{ g/mol}} \approx 0.050 \text{ moles} \]
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Calculate moles of HCl: \[ \text{Moles of HCl} = \frac{1.825 \text{ g}}{36.5 \text{ g/mol}} \approx 0.050 \text{ moles} \]
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From the balanced equation, 1 mole of ZnCO$_3$ requires 2 moles of HCl. Therefore for 0.050 moles of ZnCO$_3$, we need: \[ 0.050 \text{ moles ZnCO}_3 \times 2 = 0.100 \text{ moles HCl} \]
Since we only have 0.050 moles of HCl available, HCl is the limiting reactant, and ZnCO$_3$ is in excess.
Excess Reactant: ZnCO$_3$ (Zinc carbonate)
b. What mass of zinc chloride would you expect to make?
From the balanced equation:
- 1 mole of ZnCO$_3$ produces 1 mole of ZnCl$_2$.
Since HCl is limiting, we calculate how much ZnCl$_2$ can be produced from 0.050 moles of HCl.
As we found previously, 0.050 moles of ZnCO$_3$ would react with 0.100 moles of HCl to produce 0.050 moles of ZnCl$_2$.
Therefore, the mass of ZnCl$_2$ produced is: \[ \text{Molar mass of ZnCl}_2 = 65.4 + (35.5 \times 2) = 65.4 + 71 = 136.4 \text{ g/mol} \] \[ \text{Mass of ZnCl}_2 = 0.050 \text{ moles} \times 136.4 \text{ g/mol} = 6.82 \text{ g} \]
This gives you all the calculations done step by step and answers provided.
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