For each of the following balanced equations, calculate how many grams of each product would be produced by complete reaction of
15.0 g of the reactant
indicated in boldface. (*)
a. *2BC13(s)* + 3H2(g) -> 2B(s) + 6HCl(g)
b. *2Cu S(s)* + 302(g) → 2Cu2O(s) + 2SO2(g)
C. 2Cu2O(s) + *Cu2S(s)* -> 6Cu(s) + SO2(g)
d. CaCO3(s) + *SiO2(s)* → CaSiO3(s) + CO2(g)
indicated in boldface. (*)
a. *2BC13(s)* + 3H2(g) -> 2B(s) + 6HCl(g)
b. *2Cu S(s)* + 302(g) → 2Cu2O(s) + 2SO2(g)
C. 2Cu2O(s) + *Cu2S(s)* -> 6Cu(s) + SO2(g)
d. CaCO3(s) + *SiO2(s)* → CaSiO3(s) + CO2(g)
1 answer:
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Answer:
1. 25 moles water.
2. 41.2 grams of sodium hydroxide.
3. 0.25 grams of sugar.
4. 340.6 grams of ammonia.
5. 4.5x10²³ molecules of sulfur dioxide.
Explanation:
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In this case, since the mole-mass-particles relationships are studied by considering the Avogadro's number for the formula units and the molar mass for the mass of one mole of substance, we proceed as shown below:
1. Here, we use the Avogadro's number to obtain the moles in the given molecules of water:
2. Here, since the molar mass of NaOH is 40.00 g/mol, we obtain:
3. Here, since the molar mass of C6H12O6 is 180.15 g/mol:
4. Here, since the molar mass of ammonia is 17.03 g/mol:
5. Here, since the molar mass of SO2 is 64.06 g/mol:
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