Need help in two hours ..there are only two hours left to submit my homework 1-A
ID: 885960 • Letter: N
Question
Need help in two hours ..there are only two hours left to submit my homework
1-A mixture of 56.0 g of S and 1.00×102 g of Cl2 reacts completely to form S2Cl2 and SCl2 .
Find the mass of S2Cl2 formed.
2-Lead is found in Earth's crust as several different lead ores. Suppose a certain rock is composed of 38.0% PbS (galena), 25.0% PbCO3 (cerussite), and 17.4% PbSO4 (anglesite). The remainder of the rock is composed of substances containing no lead.
How much of this rock (in kg ) must be processed to obtain 5.5 metric tons of lead? (A metric ton is 1000 kg .)
Express your answer using two significant figures.
Explanation / Answer
A mixture of 56.0 g of S and 1.00×102 g of Cl2 reacts completely to form S2Cl2 and SCl2 .
Find the mass of S2Cl2 formed.
2 S + Cl2 = S2Cl2
moles S = 56.0 g/ 32.066 g/mol=1.75
moles Cl2 = 1.00×102 g/70.906 g/mol= 1.41
the ratio between S and Cl2 is 2 : 1
moles S required = 1.41 x 2 = 2.82
we have only 1.75 moles of S so it is the limiting reactant
moles S2Cl2 = 1.75/2=0.875
mass S2Cl2 = 0.875 x 135.038 g/mol= 118 g
2-Lead is found in Earth's crust as several different lead ores. Suppose a certain rock is composed of 38.0% PbS (galena), 25.0% PbCO3 (cerussite), and 17.4% PbSO4 (anglesite). The remainder of the rock is composed of substances containing no lead.
How much of this rock (in kg ) must be processed to obtain 5.5 metric tons of lead? (A metric ton is 1000 kg .)
at (207 g/mol Pb) / (239.3 g/mol PbS) .... PbS is 86.6 % Pb
so 86.6 % of 38.0% = 32.91 % of the rock is lead from PbS
at (207 g/mol Pb) / (303.3 g/mol PbSO4) .... PbSO4 is 68.3 % Pb
so 68.3 % of 17.4% = 11.88 % of the rock is lead from PbSO4
at (207 g/mol Pb) / (267.2 g/mol PbCO3) .... PbCO3 is 77.54 % Pb
so 77.54 % of 25.0% = 19.38 % of the rock is lead from PbCO3
19.38 + 11.88 + 32.91 = 64.2 % of the rock is Pb
soscaling it up:
5.2 metric tons of Pb (100% rock) / (64.2 % Pb) = 8.56 metric tons of rock is needed
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