Coke is an impure form of carbon that is often used in the industrial production of metals from their oxides. If a sample of coke is \(95 \%\) carbon by mass, determine the mass of coke needed to react completely with 1.0 ton of copper(II) oxide. $$2 \mathrm{CuO}(s)+\mathrm{C}(s) \longrightarrow 2 \mathrm{Cu}(s)+\mathrm{CO}_{2}(g)$$

Short Answer

Expert verified
We need approximately 79.5 kg of coke to react completely with 1.0 ton of copper(II) oxide.

Step by step solution

01

Molar mass of CuO and C

First, we must calculate the molar mass of copper(II) oxide (CuO) and carbon (C). Molar mass of Cu = 63.55 g/mol (from the periodic table) Molar mass of O = 16.00 g/mol (from the periodic table) Molar mass of C = 12.01 g/mol (from the periodic table) Molar mass of CuO = Molar mass of Cu + Molar mass of O Molar mass of CuO = 63.55 g/mol + 16.00 g/mol = 79.55 g/mol **Step 2: Calculate the moles of CuO in 1.0 ton and moles of C required for complete reaction**
02

Moles of CuO and C

We are given 1.0 ton of CuO, which we need to convert to grams: 1.0 ton = 1000 kg = 1,000,000 g Now we can calculate the moles of CuO: moles of CuO = mass of CuO / molar mass of CuO moles of CuO = 1,000,000 g / 79.55 g/mol = 12579.362 moles From the balanced chemical equation, we know that 1 mole of C reacts with 2 moles of CuO. So, to react completely with 12579.362 moles of CuO, we need: moles of C = (1/2) * moles of CuO = (1/2) * 12579.362 moles = 6289.681 moles **Step 3: Calculate the mass of carbon required for complete reaction**
03

Mass of carbon required

To find the mass of carbon required for complete reaction, we can use the molar mass of carbon and the moles of carbon required. Mass of carbon = moles of C * molar mass of C Mass of carbon = 6289.681 moles * 12.01 g/mol = 75,491.925 g **Step 4: Calculate the mass of coke needed (using the 95% carbon by mass information)**
04

Mass of coke needed

Since the coke sample is 95% carbon by mass, we need to find the total mass of coke required to supply the 75,491.925 g of carbon needed. Let x be the mass of coke needed. Then, 0.95 * x = 75,491.925 g. We can solve for x: x = 75,491.925 g / 0.95 = 79,467.290 g Hence, we need 79,467.290 g (approximately 79.5 kg) of coke to react completely with 1.0 ton of copper(II) oxide.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Natural rubidium has the average mass of 85.4678 u and is composed of isotopes \(^{85} \mathrm{Rb}\) (mass \(=84.9117 \mathrm{u}\) ) and \(^{87} \mathrm{Rb}\). The ratio of atoms \(^{85} \mathrm{Rb} /^{87} \mathrm{Rb}\) in natural rubidium is \(2.591 .\) Calculate the mass of \(^{87} \mathrm{Rb}\).

Express the composition of each of the following compounds as the mass percents of its elements. a. formaldehyde, \(\mathrm{CH}_{2} \mathrm{O}\) b. glucose, \(C_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) c. acetic acid, \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\)

A compound containing only sulfur and nitrogen is \(69.6 \% \mathrm{S}\) by mass; the molar mass is \(184 \mathrm{g} / \mathrm{mol}\). What are the empirical and molecular formulas of the compound?

Consider the following unbalanced equation: $$\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{CaSO}_{4}(s)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q)$$ What masses of calcium sulfate and phosphoric acid can be produced from the reaction of \(1.0 \mathrm{kg}\) calcium phosphate with \(1.0 \mathrm{kg}\) concentrated sulfuric acid \(\left(98 \% \mathrm{H}_{2} \mathrm{SO}_{4}\text { by mass)? }\right.\)

A sample of a hydrocarbon (a compound consisting of only carbon and hydrogen) contains \(2.59 \times 10^{23}\) atoms of hydrogen and is \(17.3 \%\) hydrogen by mass. If the molar mass of the hydrocarbon is between 55 and \(65 \mathrm{g} / \mathrm{mol}\), what amount (moles) of compound is present, and what is the mass of the sample?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free