Answer:
From the equation you will see that 1 mol of propane generates 4 mols of water.
Since the molar mass of water
M
(
H2O)=2×1+16=18g/mol
2 mol propane will generate
2
×4×18=144g of water
Explanation:
Since the molar mass of water
M
(
H2O)=2×1+16=18g/mol
2 mol propane will generate
2
×4×18=144g of water
Answer: 5.1 gram
Explanation:
To calculate the moles :
According to stoichiometry :
1 mole of require = 1 mole of
Thus 0.41 moles of will require= of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent.
Moles of HCl left = (0.55-0.41) = 0.14
Mass of left =
Thus 5.1 g of hydrochloric acid could be left over by the chemical reaction.
Answer:
a. sulfur difluoride SF₂
b. sulfur hexafluoride SF₆
c. sodium dihydrogen phosphate NaH₂PO₄
d. lithium nitride Li₃N
e. chromium(III) carbonate Cr₂(CO₃)₃
f. tin(II) fluoride SnF₂
g. ammonium acetate NH₄(CH₃COO)
h. ammonium hydrogen sulfate NH₄(HSO₄)
i. cobalt(III) nitrate Co(NO₃)₃
j. mercury(I) chloride Hg₂Cl₂
k. potassium chlorate KClO₃
l. sodium hydride NaH
Explanation:
The names give us information about the composition. First, we mention the cation and then the anion. In the formula, we follow the same order. Each part has a charge but the resulting compound is electrically neutral.
The reactants of the reaction will be solid gold (III) sulfide and hydrogen gas, while the products will be pure solid gold and hydrogen sulfide gas.
<h3>Chemical reactions</h3>
Reactants react together during reactions to arrive at products.
In this case, solid gold (III) sulfide and hydrogen gas react according to the following equation:
Reactants = solid gold (III) sulfide, hydrogen gas
Products: pure solid gold, hydrogen sulfide gas
More on chemical reactions can be found here: brainly.com/question/1689737