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A typical formulation for window glass is 75% SiO2, 15% Na2O, and 10.% CaO by mass. What masses of sand (SiO2), sodium carbonate, and calcium carbonate must be combined to produce 1.00 kg of glass after carbon dioxide is driven off by thermal decomposition of the carbonates?

Short Answer

Expert verified

The mass of sand, sodium carbonate and calcium carbonate is 0.75 kg, 0.26 kg, and 0.18 kg, respectively.

Step by step solution

01

Determination of mass of sand (SiO2)

The reaction for the decomposition of sodium carbonate is,

Na2CO3(s)→Na2O(s)+CO2(g)....................(1)

The reaction for the decomposition of calcium carbonate is,

CaCO3(s)→CaO(s)+CO2(g)...................(2)

The percentage of SiO2 in glass is 75%.

The mass of SiO2 in 1.00 kg of glass is,

=75100SiO2×1.00kg×1000g1kg=750g=0.75kgSiO2

Hence, the mass of the sand is 0.75 kg.

02

Determination of mass of sodium carbonate

The percentage of Na2O in glass is 15%.

The mass of Na2O in 1.00 kg of glass is,

=15100Na2O×1.00kg×1000g1kg=150g=0.15kgNa2O

From equation (1), it can be concluded that 1 mol of sodium carbonate produces 1 mol of Na2O.

So, the mass of sodium carbonate is,

=0.15kg×1000g1kg×1molNa2O61.98gNa2O×1molNa2CO31molNa2O×105.99Na2CO31molNa2CO3×1kg1000g=0.26kgNa2CO3

Hence, the mass of sodium carbonate is 0.26 kg.

03

Determination of mass of calcium carbonate

The percentage of CaO in glass is 10%.

The mass of Na2O in 1.00 kg of glass is,

=10100CaO×1.00kg×1000g1kg=100g=0.10kgCaO

From equation (1), it can be concluded that 1 mol of calcium carbonate produces 1 mol of CaO.

So, the mass of calcium carbonate is,

=0.10kg×1000g1kg×1molNa2O56.08gCaO×1molCaCO31molCaO×100.09CaCO31molCaCO3×1kg1000g=0.18kgCaCO3

Hence, the mass of calcium carbonate is 0.18 kg.

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Most popular questions from this chapter

On a lab exam, you have to find the concentrations of the monoprotic (one proton per molecule) acids HA and HB. You are given 43.5mL of HA solution in one flask. A second flask contains 37.2mL of HA, and you add enough HB solution to it to reach a final volume of 50.0mL. You titrate the first HA solution with 87.3mL of 0.0906M NaOH and the mixture of HA and HB in the second flask with 96.4mL of the NaOH solution. Calculate the molarity of the HA and HB solutions.

In the process of pickling, rust is removed from newly produced steel by washing the steel in hydrochloric acid:

(1)6HCl(aq)+Fe2O3(s)→2FeCl3(aq)+3H2O(l)

During the process, some iron is lost as well:

(2)2HCl(aq)+Fe(s)→FeCl2(aq)+H2(g)

(a) Which reaction, if either, is a redox process? (b) If reaction 2 did not occur and all the HCl were used, how many grams of Fe2O3 could be removed and FeCl3 produced in a 2.50x103-L bath of 3.00 MHCl? (c) If reaction 1 did not occur and all the HCl were used, how many grams of Fe could be lost and FeCl2 produced in a 2.50x103-L bath of 3.00 MHCl? (d) If 0.280 g of Fe is lost per gram of Fe2O3 removed, what is the mass ratio of FeCl2to FeCl3?

Magnesium is used in lightweight alloys for airplane bodies and other structures. The metal is obtained from seawater in a process that includes precipitation, neutralization, evaporation, and electrolysis. How many kilograms of magnesium can be obtained from 1.00 km3 of seawater if the initial Mg2+ concentration is 0.13% by mass (dof seawater = 1.04 g/mL)?

Complete the following precipitation reactions with balanced molecular, total ionic, and net ionic equations:

(a)Hg2(NO3)2(aq)+KI(aq)→(b)FeSO4(aq)+Sr(OH)2(aq)→

There are various methods for finding the composition of an alloy (a metal-like mixture). Show that calculating the mass % of Mg in a magnesium-aluminum alloy (d = 2.40 g/cm3) gives the same answer (within rounding) using each of these methods:

(a) a0.263-gsampleofalloy(dofMg=1.74g/cm3;dofAl=2.70g/cm3);

(b) an identical sample reacting with excess aqueous HCl forms 1.38x10-2molofH2;

(c) an identical sample reacting with excess O2 forms 0.483 g of oxide.

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