Question
Sulfuric acid can be prepared starting with the sulfide ore, cuprite $\left(\mathrm{Cu}_{2} \mathrm{S}\right) .$ If each $\mathrm{S}$ atom in $\mathrm{Cu}_{2} \mathrm{S}$ leads to one molecule of $\mathrm{H}_{2} \mathrm{SO}_{4},$ what mass of $\mathrm{H}_{2} \mathrm{SO}_{4}$ can be produced from $3.00 \mathrm{kg}$ of $\mathrm{Cu}_{2} \mathrm{S} ?$
Step 1
We know that 1 kg = 1000 g, so we have: \[3.00 \, \text{kg} \, \mathrm{Cu}_{2} \mathrm{S} \times \frac{1000 \, \text{g}}{1 \, \text{kg}} = 3000 \, \text{g} \, \mathrm{Cu}_{2} \mathrm{S}\] Show more…
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Sulfuric acid can be prepared starting with the sulfide ore, cuprite $\left(\mathrm{Cu}_{2} \mathrm{S}\right) .$ If each $\mathrm{S}$ atom in $\mathrm{Cu}_{2} \mathrm{S}$ leads to one molecule of $\mathrm{H}_{2} \mathrm{SO}_{4},$ what is the theoretical yield of $\mathrm{H}_{2} \mathrm{SO}_{4}$ from $3.00 \mathrm{kg}$ of $\mathrm{Cu}_{2} \mathrm{S}$ ?
Sulfuric acid can be produced from a sulfide ore such as iron pyrite by the following sequence of reactions: $$\begin{aligned} 4 \mathrm{FeS}_{2}(\mathrm{s})+11 \mathrm{O}_{2}(\mathrm{g}) & \longrightarrow 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})+8 \mathrm{SO}_{2}(\mathrm{g}) \\ 2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) & \longrightarrow 2 \mathrm{SO}_{3}(\mathrm{g}) \\ \mathrm{SO}_{3}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\ell) & \longrightarrow \mathrm{H}_{2} \mathrm{SO}_{4}(\ell) \end{aligned}$$ Starting with $525 \mathrm{kg}$ of $\mathrm{FeS}_{2}$ (and an excess of other reactants), what mass of pure $\mathrm{H}_{2} \mathrm{SO}_{4}$ can be prepared?
Sulfuric acid can be prepared starting with the sulfide ore, cuprite $\left(\mathrm{Cu}_{2} \mathrm{S}\right) .$ If each $\mathrm{S}$ atom in $\mathrm{Cu}_{2} \mathrm{S}$ leads to one molecule of $\mathrm{H}_{2} \mathrm{SO}_{4},$ what is the theoretical yield of $\mathrm{H}_{2} \mathrm{SO}_{4}$ from $3.00 \mathrm{kg}$ of $\mathrm{Cu}_{2} \mathrm{S} ?$
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