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The following equation represents the reaction of calcium disilicide \(\ce{(CaSi2)}\) with antimony trichloride \(\ce{(SbCl3)}\) to produce calcium chloride \(\ce{(CaCl2)}\), silicon \(\ce{(Si)}\), and antimony \(\ce{(Sb)}\):
\(\ce{a CaSi2 + b SbCl3 -> c Si + d Sb + e CaCl2}\)
What are the stoichiometric values for a, b, c, d, and e in the balanced equation?
\(D\)
→ Balance the chlorine atoms to begin:
\(\ce{a CaSi2 + 2SbCl3 -> c Si + d Sb + 3CaCl2}\)
→ As \(\ce{Si}\) and \(\ce{Sb}\) are by themselves on the right hand side they are be easily balanced last.
→ Next, balance the \(\ce{Ca}\) atoms on the left hand side:
\(\ce{3CaSi2 + 2SbCl3 -> c Si + d Sb + 3CaCl2}\)
→ The values for \(c\) and \(d\) can be added to balance the remaining atoms.
\(\ce{3CaSi2 + 2SbCl3 -> 6 Si + 2 Sb + 3CaCl2}\)
\(\Rightarrow D\)
Which is the correct balanced formula equation for the reaction of potassium with water?
\(C\)
→ The correct chemical equation is:
\(\ce{2K(s) + 2H2O(l) -> 2KOH(aq) + H2(g)}\)
→ The state of water is always liquid. An aqueous solution refers to a solution where a substance has been dissolved into water.
\(\Rightarrow C\)
In an experiment, calcium carbonate \(\ce{(CaCO3)}\) is heated strongly to produce calcium oxide \(\ce{(CaO)}\) and carbon dioxide according to the reaction below:
\(\ce{CaCo3(s) -> CaO(s) + CO2(g)}\)
A student starts with 50.0 g of calcium carbonate. After heating, they collect 28.0 g of calcium oxide.
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a. \(22.0\ \text{g}\)
b. The law of conservation of mass states that the mass of reactants in a chemical reaction must equal the mass of the products.
→ Here, the 50.0 g of calcium carbonate decomposes into 28.0 g of calcium oxide and 22.0 g of carbon dioxide gas.
→ The total mass of products (28.0 g + 22.0 g) equals the initial mass of reactants (50.0 g), confirming that mass is conserved in this reaction.
a. According to the law of conservation of mass, the total mass of reactants must equal the total mass of products.
Since mass must be conserved:
| \(m\ce{(CO2)}\) | \(=m\ce{(CaCO3)}-m\ce{(CaO)}\) | |
| \(= 50.0-28.0\) | ||
| \(=22.0\ \text{g}\) |
b. The law of conservation of mass states that the mass of reactants in a chemical reaction must equal the mass of the products.
→ Here, the 50.0 g of calcium carbonate decomposes into 28.0 g of calcium oxide and 22.0 g of carbon dioxide gas.
→ The total mass of products (28.0 g + 22.0 g) equals the initial mass of reactants (50.0 g), confirming that mass is conserved in this reaction.
What numbers are required to correctly balance this equation?
__\(\ce{Fe2O3 +}\) __\(\ce{CO ->}\) __\(\ce{Fe +}\) __\(\ce{CO2}\)
\(A\)
→ Start by balancing the iron \(\ce{Fe}\) atoms:
→ There are 2 Fe atoms in \(\ce{Fe2O3}\), so we place a 2 in front of \(\ce{Fe}\) on the product side.
\(\ce{Fe2O3 + CO -> 2Fe + CO2}\)
→ Next, balance the oxygen \(\ce{O}\) atoms:
→ \(\ce{Fe2O3}\) has 3 oxygen atoms, and we need 3 \(\ce{CO}\) molecules on the reactant side to balance the 3 \(\ce{CO2}\) molecules on the product side.
\(\ce{Fe2O3 + 3CO -> 2Fe + 3CO2}\)
→ Finally, check all atoms to ensure balance.
\(\Rightarrow A\)
Balance the following chemical equations:
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a. \(\ce{2HCl(aq) + Zn(s) -> ZnCl2(aq) + H2(g)}\)
b. \(\ce{2C2H6(l) + 7O2(g) -> 6H2O(l) + 4CO2(g)}\)
c. \(\ce{2H3PO4(aq) + 3CuCO3(aq) -> Cu3(PO4)2(aq) + 3CO2(g) + 3H2O(l)}\)
d. \(\ce{CuSO4(aq) + 2AgNO3(aq) -> Ag2SO4(s) + Cu(NO3)2(aq)}\)
a. \(\ce{2HCl(aq) + Zn(s) -> ZnCl2(aq) + H2(g)}\)
b. \(\ce{2C2H6(l) + 7O2(g) -> 6H2O(l) + 4CO2(g)}\)
c. \(\ce{2H3PO4(aq) + 3CuCO3(aq) -> Cu3(PO4)2(aq) + 3CO2(g) + 3H2O(l)}\)
d. \(\ce{CuSO4(aq) + 2AgNO3(aq) -> Ag2SO4(s) + Cu(NO3)2(aq)}\)